Straight Lines

If the pair of straight lines $x^2 - 2 kxy - y^2 = 0$ bisect the angle between the pair of straight lines $x^2 - 2 \ell x y - y^2$, show that the later pair also bisects the angle between the former.

Comparing the equation $\;\;$ $x^2 - 2 \ell x y - y^2$ $\;\;\; \cdots \; (1)$

with the standard equation $\;\;$ $ax^2 = 2hxy + by^2 = 0$ $\;\;\; \cdots \; (2)$ $\;$

gives $\;\;$ $a = 1$, $\;\;\;$ $h = - \ell$, $\;\;\;$ $b = -1$

Equation of angle bisector of pair of lines given by equation $(2)$ is

$\dfrac{x^2 - y^2}{ab} = \dfrac{xy}{h}$ $\;\;\; \cdots \; (3)$

Substituting the values of $a$, $b$ and $h$ in equation $(3)$, the equation of angle bisector of the pair of lines given by equation $(1)$ is

$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{- \ell}$

i.e. $\ell x^2 - \ell y^2 = xy$ $\implies$ $\ell x^2 - xy - \ell y^2 = 0$ $\;\;\; \cdots \; (4)$

Given: The pair of straight lines $x^2 - 2kxy - y^2 = 0$ $\;\;\; \cdots \; (5)$

is the angle bisector the pair of lines given by equation $(1)$.

Since equations $(4)$ and $(5)$ represent the same angle bisector, comparing the like terms in both equations gives

$\dfrac{\ell}{1} = \dfrac{1}{2k} = \dfrac{\ell}{1}$ $\implies$ $k = \dfrac{1}{2 \ell}$ $\;\;\; \cdots \; (6)$

Comparing equation $(5)$ with the standard equation $\;\;$ $Ax^2 + 2 Hxy + By^2 = 0$

gives $\;\;$ $A = 1$, $\;\;\;$ $H = -k$, $\;\;\;$ $B = -1$

Equation of angle bisector of pair of lines given by equation $(5)$ is

$\dfrac{x^2 - y^2}{1 \times \left(-1\right)} = \dfrac{xy}{-k}$

i.e. $kx^2 - ky^2 = xy$ $\implies$ $k x^2 - xy - ky^2 = 0$ $\;\;\; \cdots \; (7)$

Substituting the value of $k$ from equation $(6)$ in equation $(7)$ gives

$\dfrac{x^2}{2 \ell} - xy - \dfrac{y^2}{2 \ell} = 0$

i.e. $x^2 - 2 \ell xy - y^2 = 0$ $\;\;\;$ which are pair of lines given by equation $(1)$.

$\therefore$ $\;$ The pair of equations given by $(1)$ bisect the angle between the pair of lines given by equation $(5)$.

Hence proved.

Straight Lines

Prove that one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$

Let $m_1$ and $m_2$ be the slopes of the lines given by the equation $\;\;$ $ax^2 + 2hxy + by^2 = 0$ $\;\;\; \cdots \; (1)$

The two lines given by equation $(1)$ are

$y - m_1 x = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - m_2 x = 0$ $\;\;\; \cdots \; (2b)$

Now, $\;$ $m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (3b)$

Let the line given by equation $(2a)$ bisect the coordinate axes.

Then the angle between the line and the coordinate axis (say, X axis) is $\dfrac{\pi}{4}$

$\therefore$ $\;$ Slope of line given by equation $(2a)$ is $m_1 = \tan \left(\dfrac{\pi}{4}\right) = 1$

Substituting $m_1 = 1$ in equation $(3a)$ gives

$1 + m_2 = \dfrac{-2h}{b}$ $\implies$ $m_2 = \dfrac{-2h}{b} - 1$ $\;\;\; \cdots \; (4a)$

Substituting $m_1 = 1$ in equation $(3b)$ gives

$m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ We have from equations $(4a)$ and $(4b)$,

$\dfrac{-2h}{b} - 1 = \dfrac{a}{b}$

i.e. $- 2h - b = a$ $\implies$ $a + b = - 2h$ $\implies$ $\left(a + b\right)^2 = 4 h^2$

Hence, one of the straight lines given by $ax^2 + 2hxy + by^2 = 0$ will bisect the angle between the coordinate axes if $\left(a + b\right)^2 = 4h^2$.

Straight Lines

Show that the equation $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ represents a pair of parallel lines. Find the distance between them.

Comparing the given equation $\;$ $4x^2 + 4 xy + y^2 - 6x - 3y - 4 = 0$ $\;\;\; \cdots \; (1)$

with the standard equation $\;$ $ax^2 + 2 hxy + by^2 + 2 gx + 2 fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$\begin{aligned} a = 4, & \hspace{1em} 2 f = -3 \implies f = \dfrac{-3}{2} \\\\ b = 1, & \hspace{1em} 2 g = -6 \implies g = -3 \\\\ c = -4, & \hspace{1em} 2 h = 4 \implies h = 2 \end{aligned}$

Two straight lines represented by the standard equation $(2)$ are parallel if $\;\;$ $af^2 = bg^2$

Now, $a f^2 = 4 \times \left(\dfrac{-3}{2}\right)^2 = 9$ $\;\;\; \cdots \; (3a)$

and $bg^2 = 1 \times \left(-3\right)^2 = 9$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$ $\;$ $af^2 = bg^2$

$\implies$ Equation $(1)$ represents a pair of parallel lines.

Distance between the pair of parallel lines $= d = 2 \times \sqrt{\dfrac{g^2 - ac}{a \left(a + b\right)}}$

i.e. $d = 2 \times \sqrt{\dfrac{\left(-3\right)^2 - 4 \times \left(-4\right)}{4 \left(4 + 1\right)}} = 2 \times \sqrt{\dfrac{9 + 16}{4 \times 5}} = \sqrt{5}$

$\therefore$ $\;$ Distance between the pair of parallel lines represented by equation $(1)$ is $d = \sqrt{5}$ units.

Straight Lines

A $\triangle OPQ$ is formed by the pair of straight lines $x^2 - 4 xy + y^2 = 0$ and the line $PQ$. The equation of $PQ$ is $x + y - 2 = 0$. Find the equation of the median of $\triangle OPQ$ drawn from the origin.

$x^2 - 4xy + y^2 = 0$ $\;\;\; \cdots \; (1)$

represents a pair of straight lines passing through the origin $O \left(0,0\right)$.

Equation $(1)$ can be written as

$x^2 - 4xy + 4y^2 - 3y^2 = 0$

i.e. $\left(x - 2y\right)^2 - \left(\sqrt{3}y\right)^2 = 0$

i.e. $\left(x - 2y + \sqrt{3}y\right) \left(x - 2y - \sqrt{3}y\right) = 0$

$\therefore$ $\;$ Equation $(1)$ represents the pair of lines

$x - y \left(2 - \sqrt{3}\right) = 0$ and $x - y \left(2 + \sqrt{3}\right) = 0$

i.e. $x = y \left(2 - \sqrt{3}\right)$ $\;\;\; \cdots \; (2a)$ and

$x = y \left(2 + \sqrt{3}\right)$ $\;\;\; \cdots \; (2b)$

Equation of line $PQ$ is $\;$ $x + y - 2 = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(2a)$ and $(3)$ simultaneously we have,

$y \left(2 - \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 - \sqrt{3}}$ $\;\;\; \cdots \; (4a)$

Substituting the value of $y$ from equation $(4a)$ in equation $(2a)$ we get,

$x = \left(\dfrac{2}{3 - \sqrt{3}}\right) \times \left(2 - \sqrt{3}\right)$

i.e. $x = \dfrac{2 \times \left(2 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}{\left(3 - \sqrt{3}\right) \times \left(3 + \sqrt{3}\right)}$

i.e. $x = \dfrac{2 \times \left(6 + 2 \sqrt{3} - 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 - \sqrt{3}}{3}$

$\therefore$ $\;$ The point of intersection of equations $(2a)$ and $(3)$ is $A \left(\dfrac{3 - \sqrt{3}}{3}, \dfrac{2}{3 - \sqrt{3}}\right)$

Solving equations $(2b)$ and $(3)$ simultaneously we have,

$y \left(2 + \sqrt{3}\right) + y = 2$ $\implies$ $y = \dfrac{2}{3 + \sqrt{3}}$ $\;\;\; \cdots \; (4b)$

Substituting the value of $y$ from equation $(4b)$ in equation $(2b)$ we get,

$x = \left(\dfrac{2}{3 + \sqrt{3}}\right) \times \left(2 + \sqrt{3}\right)$

i.e. $x = \dfrac{2 \times \left(2 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}{\left(3 + \sqrt{3}\right) \times \left(3 - \sqrt{3}\right)}$

i.e. $x = \dfrac{2 \times \left(6 - 2 \sqrt{3} + 3 \sqrt{3} - 3\right)}{9 - 3}$ $\implies$ $x = \dfrac{3 + \sqrt{3}}{3}$

$\therefore$ $\;$ The point of intersection of equations $(2b)$ and $(3)$ is $B \left(\dfrac{3 + \sqrt{3}}{3}, \dfrac{2}{3 + \sqrt{3}}\right)$

Midpoint of $AB$ is $\;$ $M = \left(\dfrac{\dfrac{3 - \sqrt{3} + 3 + \sqrt{3}}{3}}{2}, \dfrac{\dfrac{2}{3 - \sqrt{3}} + \dfrac{2}{3 + \sqrt{3}}}{2}\right)$

i.e. $M = \left(1, \dfrac{6 + 2 \sqrt{3} + 6 - 2 \sqrt{3}}{2 \times \left(3 - \sqrt{3}\right) \left(3 + \sqrt{3}\right)}\right)$

i.e. $M = \left(1, \dfrac{12}{2 \times 6}\right)$ $\implies$ $M = \left(1,1\right)$

$\therefore$ $\;$ Equation of median $MO$ is

$y - 0 = \left(\dfrac{1 - 0}{1 - 0}\right) \left(x - 0\right)$

$\therefore$ $\;$ The required equation of the median of $\triangle OPQ$ drawn from the origin is $\;$ $y = x$

Straight Lines

Find $\;$ $p$ $\;$ and $\;$ $q$ $\;$ if the equation $\;$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;$ represents a pair of perpendicular lines.

Comparing the equation $\hspace{1em}$ $6x^2 + 5xy - py^2 + 7x + qy - 5 = 0$ $\;\;\; \cdots \; (1)$

with the standard equation $\hspace{1em}$ $ax^2 + 2 hxy + by^2 + 2gx + 2 fy + c = 0$ $\;$ gives

$\begin{aligned} a = 6, & \hspace{1em} 2 f = q \implies f = \dfrac{q}{2} \\\\ b = - p, & \hspace{1em} 2 g = 7 \implies g = \dfrac{7}{2} \\\\ c = -5, & \hspace{1em} 2 h = 5 \implies h = \dfrac{5}{2} \end{aligned}$

Given: The two lines represented by equation $(1)$ are perpendicular to each other.

Now, a pair of lines are perpendicular when $\;\;\;$ $a + b = 0$

Substituting the values of $a$ and $b$ we have,

$6 - p = 0$ $\implies$ $p = 6$

Condition for equation $(1)$ to represent a pair of lines is: $\hspace{1em}$ $abc + 2 fgh - af^2 - bg^2 - ch^2 = 0$

Substituting the values of $a$, $b$, $c$, $f$, $g$ and $h$ we have,

$6 \times \left(-6\right) \times \left(-5\right) + 2 \times \left(\dfrac{q}{2}\right) \times \left(\dfrac{7}{2}\right) \times \left(\dfrac{5}{2}\right)$
$\hspace{3em}$ $- 6 \times \left(\dfrac{q}{2}\right)^2 - \left(-6\right) \times \left(\dfrac{7}{2}\right)^2 - \left(-5\right) \times \left(\dfrac{5}{2}\right)^2 = 0$

i.e. $180 + \dfrac{35q}{4} - \dfrac{6q^2}{4} + \dfrac{294}{4} + \dfrac{125}{4} = 0$

i.e. $6 q^2 - 35 q - 1139 = 0$

i.e. $q = \dfrac{35 \pm \sqrt{35^2 + 4 \times 6 \times 1139}}{2 \times 6}$

i.e. $q = \dfrac{35 \pm \sqrt{28561}}{12} = \dfrac{35 \pm 169}{12}$

$\implies$ $q = 17$ $\;$ or $\;$ $q = \dfrac{-67}{6}$

Straight Lines

If the slope of one of the straight lines $ax^2 + 2hxy + by^2 = 0$ is three times the other, then show that $3h^2 = 4ab$.

Let $m_1$ and $m_2$ be the slopes of the two lines.

Given: $\hspace{1em}$ $m_1 = 3m_2$ $\;\;\; \cdots \; (1)$

For a pair of lines,

$m_1 + m_2 = \dfrac{-2h}{b}$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $m_1 m_2 = \dfrac{a}{b}$ $\;\;\; \cdots \; (2b)$

$\begin{aligned} \text{Now, } \left(m_1 - m_2\right)^2 & = \left(m_1 + m_2\right)^2 - 4 m_1 m_2 \\\\ & = \dfrac{4h^2}{b^2} - \dfrac{4a}{b} \hspace{1em} \left[\text{from equations }(2a) \text{ and } (2b)\right] \\\\ & = \dfrac{4 \left(h^2 - ab\right)}{b^2} \end{aligned}$

i.e. $m_1 - m_2 = \pm \dfrac{2 \sqrt{h^2 - ab}}{b}$

Consider the case $\hspace{1em}$ $m_1 - m_2 = \dfrac{2 \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (2c)$

Adding equations $(2a)$ and $(2c)$ gives

$2 m_1 = \dfrac{-2h + 2 \sqrt{h^2 - ab}}{b}$ $\implies$ $m_1 = \dfrac{-h + \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3a)$

Substituting the value of $m_1$ from equation $(3a)$ in equation $(2a)$ gives

$m_2 = \dfrac{-2h}{b} + \dfrac{h - \sqrt{h^2 -ab}}{b}$ $\implies$ $m_2 = \dfrac{-h - \sqrt{h^2 - ab}}{b}$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ In view of equations $(1)$, $(3a)$ and $(3b)$ we have

$\dfrac{-h + \sqrt{h^2 - ab}}{b} = - 3 \times \left(\dfrac{h + \sqrt{h^2 - ab}}{b}\right)$

i.e. $- h + \sqrt{h^2 - ab} = - 3h - 3 \sqrt{h^2 - ab}$

i.e. $4 \sqrt{h^2 - ab} = - 2h$ $\implies$ $2\sqrt{h^2 - ab} = - h$ $\;\;\; \cdots \; (4)$

Squaring both sides of equation $(4)$ we have,

$4 \left(h^2 - ab\right) = h^2$

i.e. $4 h^2 - 4ab = h^2$ $\implies$ $3h^2 = 4ab$ $\;\;$ Hence proved.

Note:
The same result can be obtained by taking $\;$ $m_1 - m_2 = \dfrac{-2 \sqrt{h^2 - ab}}{b}$ $\;$ in equation $(2c)$.
The values of $m_1$ and $m_2$ will be interchanged.

Straight Lines

Find the equation of the pair of straight lines passing through the point $\left(1,3\right)$ and perpendicular to the lines $2x - 3y + 1 = 0$ and $5x + y - 3 = 0$

The given lines are

$2x - 3y + 1 = 0$ $\;$ i.e. $y = \dfrac{2}{3}x + \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$

and $5x + y - 3 = 0$ $\;$ i.e. $y = - 5 x + 3$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ Slope of line given by equation $(1)$ is $= m_1 = \dfrac{2}{3}$

and slope of line given by equation $(2)$ is $= m_2 = -5$

Let the slopes of the required pair of lines be $m_3$ and $m_4$.

Since the required lines are perpendicular to the given lines, their slopes are

$m_3 = \dfrac{-1}{m_1} = \dfrac{-3}{2}$ $\;$ and $\;$ $m_4 = \dfrac{-1}{m_2} = \dfrac{1}{5}$

Now, the required lines pass through the point $\left(1,3\right)$.

$\therefore$ $\;$ The equations of the individual lines are

$\left(y - 3\right) = \dfrac{-3}{2} \left(x - 1\right)$ $\;$ and $\;$ $\left(y - 3\right) = \dfrac{1}{5} \left(x - 1\right)$

i.e. $2y - 6 = - 3x + 3$ $\;$ and $\;$ $5y - 15 = x - 1$

i.e. $3x + 2 y - 9 = 0$ $\;$ and $\;$ $x - 5y + 14 = 0$

$\therefore$ $\;$ The combined equation of the required lines is

$\left(3x + 2y - 9\right) \left(x - 5y + 14\right) = 0$

i.e. $3x^2 - 15xy + 42x + 2 xy - 10 y^2 + 28 y - 9x + 45 y - 126 = 0$

i.e. $3x^2 - 13xy - 10y^2 + 33x + 73y - 126 = 0$ $\;\;\; \cdots \; (3)$

Equation $(3)$ gives the required equation of the pair of straight lines.

Straight Lines

Show that the equation $2x^2 - xy - 3y^2 - 6x + 19 y - 20 = 0$ represents a pair of intersecting lines. Find the equations of the two lines and the angle between them.

Comparing the given equation: $\hspace{1em}$ $2x^2 - xy - 3y^2 - 6x + 19 y - 20 = 0$ $\;\;\; \cdots \; (1)$

with the standard equation: $\hspace{1em}$ $ax^2 + 2 hxy + by^2 + 2 gx + 2fy + c = 0$ $\;$ gives

$a = 2$, $\;\;\;$ $2h = -1 \implies h = \dfrac{-1}{2}$, $\;\;\;$ $b = -3$,

$2 g = - 6 \implies g = -3$, $\;\;\;$ $2f = 19 \implies f = \dfrac{19}{2}$, $\;\;\;$ $c = -20$

Condition for equation $(1)$ to represent a pair of straight lines is if

$abc + 2 fgh - af^2 - bg^2 - ch^2 = 0$

Now, $abc + 2 fgh - af^2 - bg^2 - ch^2$

$= 2 \times \left(-3\right) \times \left(-20\right) + 2 \times \left(\dfrac{19}{2}\right) \times \left(-3\right) \times \left(\dfrac{-1}{2}\right)$
$\hspace{2em}$ $- 2 \times \left(\dfrac{19}{2}\right)^2 - \left(-3\right) \times \left(-3\right)^2 - \left(-20\right) \times \left(\dfrac{-1}{2}\right)^2$

$= 120 + \dfrac{57}{2} - \dfrac{361}{2} + 27 + 5$

$= 152 - \dfrac{304}{2} = 0 $

$\therefore$ $\;$ We have $\;$ $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

$\implies$ Equation $(1)$ represents a pair of straight lines.

Factorizing the second degree terms in equation $(1)$, we have

$\begin{aligned} 2x^2 - xy - 3y^2 & \equiv 2x^2 + 2xy - 3xy - 3y^2 \\\\ & \equiv \left(2x - 3y\right) \left(x+y\right) \end{aligned}$

$\therefore$ $\;$ $2x^2 - xy - 3y^2 - 6x + 19 y - 20 \equiv \left(2x - 3y + c_1\right) \left(x + y + c_2\right)$ $\;\;\; \cdots \; (2)$

where $c_1$ and $c_2$ are constants.

Equating the coefficients of the $x$ terms in equation $(2)$ we get,

$-6 = c_1 + 2c_2$ $\;\;\; \cdots \; (3a)$

Equating the coefficients of the $y$ terms in equation $(2)$ we get,

$19 = c_1 - 3c_2$ $\;\;\; \cdots \; (3b)$

Subtracting equations $(3a)$ and $(3b)$ we get,

$25 = -5 c_2$ $\implies$ $c_2 = -5$

Substituting the value of $c_2$ in equation $(3a)$ gives

$-6 = -10 + c_1$ $\implies$ $c_1 = 4$

$\therefore$ $\;$ The separate equations of the lines are

$2x - 3y + 4 = 0$ and $x + y - 5 = 0$

Angle between the lines $= \tan \theta = \left|\dfrac{2 \sqrt{h^2 -ab}}{a+b}\right|$

i.e. $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-1}{2}\right)^2 - 2 \times \left(-3\right)}}{2-3}\right|$

i.e. $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{1}{4}+ 6}}{-1}\right| = 2 \times \sqrt{\dfrac{25}{4}} = 5$

$\implies$ $\theta = \tan^{-1} \left(5\right)$

Straight Lines

Find the image of the point $\left(-2,3\right)$ about the line $x + 2y - 9 = 0$

Let $P \left(-2.3\right)$ be the given point.

Perpendicular distance of point P from the given line $x + 2y - 9 = 0$ is

$d = \left|\dfrac{1 \times \left(-2\right) + 2 \times 3 - 9}{\sqrt{\left(1\right)^2 + \left(2\right)^2}}\right|$

i.e. $d = \left|\dfrac{-2 + 6 - 9}{\sqrt{5}}\right| = \sqrt{5}$ units $\;\;\; \cdots \; (1)$

Let $Q \left(a, b\right)$ be the image of the point $P$ in the given line.

Then, perpendicular distance of $Q$ from the given line is $= d = \sqrt{5}$ units

i.e. $\left|\dfrac{1 \times a + 2 \times b - 9}{\sqrt{\left(1\right)^2 + \left(2\right)^2}}\right| = \sqrt{5}$

i.e. $\dfrac{a + 2b - 9}{\sqrt{5}} = \sqrt{5}$

i.e. $a + 2b - 9 = 5$ $\implies$ $a = 14 - 2b$ $\;\;\; \cdots \; (2)$

Now, distance $PQ = \sqrt{5} + \sqrt{5} = 2 \sqrt{5}$ units

Also, $PQ = \sqrt{\left(a + 2\right)^2 + \left(b - 3\right)^2}$

$\therefore$ $\;$ We have $\sqrt{\left(a + 2\right)^2 + \left(b - 3\right)^2} = 2 \sqrt{5}$

i.e. $\left(a + 2\right)^2 + \left(b - 3\right)^2 = 20$ $\;\;\; \cdots \; (3)$

Substituting the value of $a$ from equation $(2)$ in equation $(3)$ gives

$\left(14 - 2b + 2\right)^2 + \left(b - 3\right)^2 = 20$

i.e. $\left(16 - 2b\right)^2 + \left(b - 3\right)^2 = 20$

i.e. $256 - 64 b + 4 b^2 + b^2 - 6b + 9 = 20$

i.e. $5b^2 - 70b + 245 = 0$

i.e. $b^2 - 14 b + 49 = 0$

i.e. $\left(b - 7\right)^2 = 0$ $\implies$ $b - 7 = 0$ $\implies$ $b = 7$ $\;\;\; \cdots \; (4)$

Substituting the value of $b$ from equation $(4)$ in equation $(2)$ gives

$a = 14 - 2 \times 7 = 0$

$\therefore$ $\;$ The image of the point $\left(-2,3\right)$ about the given line is $\left(0,7\right)$

Straight Lines

A line is drawn perpendicular to $5x = y + 7$. Find the equation of the line if the area of the triangle formed by this line with the coordinate axes is 10 square units.

Equation of the given line can be written as: $\hspace{1em}$ $y = 5x - 7$

$\therefore$ $\;$ Slope of the given line $= m = 5$

$\because$ $\;$ The required line is perpendicular to the given line,

slope of the required line $= m_1 = \dfrac{-1}{m} = \dfrac{-1}{5}$ $\;\;\; \cdots \; (1)$

Let AB be the required line where $A \left(0,a\right)$ and $B \left(b,0\right)$.

The equation of the required line can be written the form $y - y_1 = m_1 \left(x - x_1\right)$ as

$y - 0 = \dfrac{-1}{5} \left(x - b\right)$

i.e. $5y = -x + b$ $\implies$ $b = x + 5y$ $\;\;\; \cdots \; (2)$

The equation of the required line can also be written as

$y - a = \dfrac{-1}{5} \left(x - 0\right)$

i.e. $5 y - 5 a = - x$ $\implies$ $x + 5 y = 5a$ $\;\;\; \cdots \; (3)$

Given: The required line makes a triangle $\left(\triangle AOB\right)$ of area 10 square units with the coordinate axes.

Now, area of $\triangle AOB = \dfrac{1}{2}ab$

$\therefore$ $\;$ We have $\hspace{1em}$ $\dfrac{1}{2}ab = 10$ $\implies$ $a = \dfrac{20}{b}$ $\;\;\; \cdots \; (4)$

Substituting the value of $a$ from equation $(4)$ in equation $(3)$ gives

$x + 5y = 5 \times \dfrac{20}{b}$

$\implies$ $b = \dfrac{100}{x + 5y}$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ We have from equations $(5)$ and $(2)$,

$x + 5y = \dfrac{100}{x + 5y}$

i.e. $\left(x + 5y\right)^2 = 100$

$\implies$ $x + 5 y = \pm 10$ $\;\;\; \cdots \; (6)$

Equation $(6)$ gives the equations of the required lines.

Straight Lines

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec \theta + y \text{cosec } \theta = 2a$ and $x \cos \theta - y \sin \theta = a \cos 2 \theta$, then prove that $p^2_1+p^2_2 = a^2$.

The given equations can be written as

$x \sec \theta + y \text{cosec } \theta - 2a = 0$ $\;\;\; \cdots \; (1)$

and $x \cos \theta - y \sin \theta - a \cos 2 \theta = 0$ $\;\;\; \cdots \; (2)$

Length of perpendicular from $\left(0,0\right)$ to equation $(1)$ is

$p_1 = \left|\dfrac{-2a}{\sqrt{\sec^2 \theta + \text{cosec}^2 \theta}}\right|$

i.e. $p^2_1 = \dfrac{4a^2}{\dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta}}$

i.e. $p^2_1 = \dfrac{4a^2 \times \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$

i.e. $p^2_1 = 4 a^2 \sin^2 \theta \cos^2 \theta$ $\;\;\; \cdots \; (3)$

Length of perpendicular from $\left(0,0\right)$ to equation $(2)$ is

$p_2 = \left|\dfrac{- a \cos 2 \theta}{\sqrt{\left(\cos \theta\right)^2 + \left(- \sin \theta\right)^2}}\right|$

i.e. $p^2_2 = a^2 \cos^2 2 \theta$

i.e. $p^2_2 = a^2 \left(\cos^2 \theta - \sin^2 \theta\right)^2$

i.e. $p^2_2 = a^2 \left(\cos^4 \theta + \sin^4 \theta - 2 \cos^2 \theta \sin^2 \theta\right)$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ From equations $(3)$ and $(4)$,

$\begin{aligned} p^2_1 + p^2_2 & = 4 a^2 \sin^2 \theta \cos^2 \theta + a^2 \cos^4 \theta + a^2 \sin^4 \theta - 2a^2 \cos^2 \theta \sin^2 \theta \\\\ & = a^2 \left(\cos^4 \theta + 2a^2 \cos^2 \theta \sin^2 \theta + \sin^4 \theta\right) \\\\ & = a^2 \left(\cos^2 \theta + \sin^2 \theta\right)^2 \\\\ & = a^2 \end{aligned}$

i.e. $p^2_1 + p^2_2 = a^2$ $\;\;\;$ Hence proved

Straight Lines

Find the equation of a straight line parallel to $2x + 3y = 10$ and which is such that the sum of its intercepts on the axes is 15.

Equation of the given line is: $\hspace{1em}$ $2x + 3y = 10$

i.e. $y = - \dfrac{2}{3} x + \dfrac{10}{3}$

$\therefore$ $\;$ Slope of the given line $= m = - \dfrac{2}{3}$

Since the required line is parallel to the given line,

$\therefore$ $\;$ slope of the required line $= m = - \dfrac{2}{3}$ $\;\;\; \cdots \; (1)$

Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (2)$

where $a$ and $b$ are the intercepts on the X and the Y axes respectively.

Given: Sum of intercepts on the axes $= a + b = 15$

i.e. $a = 15 - b$ $\;\;\; \cdots \; (3)$

Substituting the value of $a$ from equation $(3)$ in equation $(2)$ gives

$\dfrac{x}{15 - b} + \dfrac{y}{b} = 1$

i.e. $b \; x + \left(15 - b\right)y = b \left(15 - b\right)$

i.e. $y = \left(\dfrac{-b}{15 - b}\right) \; x + b$

$\therefore$ $\;$ Slope of the required line is $= \dfrac{b}{b - 15}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ We have from equations $(1)$ and $(4)$,

$\dfrac{b}{b - 15} = - \dfrac{2}{3}$

i.e. $3 b = - 2 b + 30$ $\implies$ $b = 6$ $\;\;\; \cdots \; (5)$

Substituting the value of $b$ from equation $(5)$ in equation $(3)$ we have,

$a = 15 - 6 = 9$ $\;\;\; \cdots \; (6)$

$\therefore$ The required equation of line is [from equation $(2)$]:

$\dfrac{x}{9} + \dfrac{y}{6} = 1$

i.e. $2x + 3y - 18 = 0$

Straight Lines

Find the equations of the two straight lines which are parallel to the line $12 x + 5 y + 2 = 0$ and at an unit distance from the point $\left(1, -1\right)$.

Equation of the given line is: $\hspace{1em}$ $12x + 5y + 2 = 0$

i.e. $y = -\dfrac{12}{5}x - \dfrac{2}{5}$

$\therefore$ $\;$ Slope of the given line is $= m = - \dfrac{12}{5}$

Since the required line is parallel to the given line,

$\therefore$ $\;$ Slope of the required line $= m = - \dfrac{12}{5}$ $\;\;\; \cdots \; (1)$

Let the equation of the required line be $\hspace{1em}$ $y = mx + c$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(1)$, equation $(2)$ becomes

$y = - \dfrac{12}{5} x + c$

i.e. $5y = -12 x + 5 c$ $\implies$ $12x + 5y - 5c = 0$ $\;\;\; \cdots \; (2a)$

Distance of a line $ax + by + c = 0$ from a point $\left(x_1, y_1\right)$ is $d = \left|\dfrac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\right|$

Given: $\left(x_1 , y_1\right) = \left(1, -1\right)$

and the distance of the line from the point $= d = 1$

$\therefore$ $\;$ We have for the required line

$1 = \left|\dfrac{12 \times 1 + 5 \times \left(-1\right) - 5 c}{\sqrt{12^2 + 5^2}}\right|$

i.e. $\pm 1 = \dfrac{7 - 5 c}{13}$

$\implies$ $13 = 7 - 5c$ $\;\;\;$ i.e. $- 5 c = 6$ $\implies$ $c = - \dfrac{6}{5}$ $\;\;\; \cdots \; (3a)$

or, $-13 = 7 - 5c$ $\;\;\;$ i.e. $- 5 c = -20$ $\implies$ $c = 4$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ In view of equation $(3a)$ equation $(2a)$ becomes

$12 x + 5 y - 5 \times \left(- \dfrac{6}{5}\right) = 0$ $\implies$ $12 x + 5 y + 6 = 0$

and in view of equation $(3b)$ equation $(2a)$ becomes

$12 x + 5 y - 5 \times 4 = 0$ $\implies$ $12 x + 5y - 20 = 0$

Therefore, the equations of the required lines are: $\hspace{1em}$ $12x + 5y + 6 = 0$ and $12 x + 5y - 20 = 0$

Straight Lines

Find the equation of the straight line parallel to $5x - 4y + 3 = 0$ and having X intercept 3.

Equation of the given line is: $\hspace{1em}$ $5x - 4 y + 3 = 0$

i.e. $y = \dfrac{5}{4}x + \dfrac{3}{4}$

$\therefore$ $\;$ Slope of the given line $= m = \dfrac{5}{4}$

Since the required line is parallel to the given line,

slope of required line $= m = \dfrac{5}{4}$

Let the intercept made by the required line on the X axis be $a$.

Given: $a = 3$

Let the required line make an angle $\alpha$ with the X axis.

Then, $\tan \alpha = \dfrac{PB}{PA} = \dfrac{y}{x - a}$

But $\tan \alpha$ gives the slope of the line.

$\therefore$ $\;$ Slope of the required line $ = m = \dfrac{y}{x - a}$

i.e. $y = m \left(x - a\right)$ $\;\;\; \cdots \; (1)$

Substituting the values of $m$ and $a$ in equation $(1)$, we have

$y = \dfrac{5}{4} \left(x - 3\right)$

i.e. $4y = 5x - 15$

or, $5x - 4y - 15 = 0$

which is the equation of the required line.

Straight Lines

A straight line is passing through the point $A \left(1,2\right)$ with slope $\dfrac{5}{12}$. Find the points on the line which are 13 units away from $A$.

The required line passes through $A \left(1,2\right)$ with slope $\dfrac{5}{12}$

$\therefore$ $\;$ Equation of the required line is: $\hspace{1em}$ $\left(y - 2\right) = \dfrac{5}{12} \left(x - 1\right)$

i.e. $12 y - 24 = 5 x - 5$

i.e. $5x - 12 y = - 19$ $\;\;\; \cdots \; (1)$

Let $P \left(p , q\right)$ be a point on the line given by equation $(1)$.

Then we have $\hspace{1em}$ $5 p - 12 q = - 19$ $\;\;\; \cdots \; (2)$

As per question, distance $AP = 13$

Now, distance $AP = \sqrt{\left(p - 1\right)^2 + \left(q - 2\right)^2}$

$\therefore$ $\;$ We have $\hspace{1em}$ $\sqrt{\left(p - 1\right)^2 + \left(q - 2\right)^2} = 13$

i.e. $\left(p - 1\right)^2 + \left(q - 2\right)^2 = 169$ $\;\;\; \cdots \; (3)$

We have from equation $(2)$, $p = \dfrac{12 q - 19}{5}$ $\;\;\; \cdots \; (4)$

Substituting the value of $p$ from equation $(4)$ in equation $(3)$ we get,

$\left(\dfrac{12 q - 19}{5} - 1\right)^2 + \left(q - 2\right)^2 = 169$

i.e. $\left(\dfrac{12 q - 24}{5}\right)^2 + \left(q - 2\right)^2 = 169$

i.e. $\dfrac{144}{25} \left(q - 2\right)^2 + \left(q - 2\right)^2 = 169$

i.e. $\dfrac{169}{25} \left(q - 2\right)^2 = 169$

i.e. $\left(q - 2\right)^2 = 25$ $\implies$ $q - 2 = \pm 5$

i.e. $q = 7$ or $q = - 3$

Substituting the values of $q$ in equation $(4)$ we get,

when $q = 7$, $p = \dfrac{12 \times 7 - 19}{5} = 13$ and

when $q = - 3$, $p = \dfrac{12 \times \left(-3\right) - 19}{5} = -11$

$\therefore$ $\;$ The required points on the line are $\left(13, 7\right)$ and $\left(-11, -3\right)$.

Straight Lines

Find the equations of straight lines passing through $\left(8,3\right)$ and having intercepts whose sum is 1.

Let the intercept on the X axis be $= a$

and the intercept on the Y axis be $= b$.

Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (1)$

Given: $\hspace{1em}$ $a + b = 1$ $\implies$ $a = 1 - b$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{x}{1 - b} + \dfrac{y}{b} = 1$ $\;\;\; \cdots \; (3)$

Now, equation $(3)$ passes through $\left(8,3\right)$.

$\therefore$ $\;$ We have: $\hspace{1em}$ $\dfrac{8}{1 - b}+ \dfrac{3}{b} = 1$

i.e. $8 b + 3 - 3b = b \left(1 - b\right)$

i.e. $5 b + 3 = b - b^2$

i.e. $b^2 + 4 b + 3 = 0$

i.e. $\left(b + 3\right) \left(b + 1\right) = 0$

$\implies$ $b = - 3$ or $b = -1$

Substituting the value of $b$ in equation $(2)$ gives

when $b = - 3$, $a = 1 - \left(-3\right) = 4$

when $b = -1$, $a = 1 - \left(-1\right) = 2$

$\therefore$ $\;$ Substituting the values of $a$ and $b$ in equation $(1)$ gives the equations of lines as

$\dfrac{x}{4} + \dfrac{y}{-3} = 1$ $\;\;\;$ i.e. $- 3x + 4y = -12$ $\;\;\;$ i.e. $\;$ $3x - 4y = 12$

and $\dfrac{x}{2} + \dfrac{y}{-1} = 1$ $\;\;\;$ i.e. $\;$ $- x + 2y = -2$ $\;\;\;$ i.e. $\;$ $x - 2y = 2$

Straight Lines

Find the equation of the line passing through the point $\left(1,5\right)$ and which divides the coordinate axes in the ratio $3 : 10$.

Let the required line cut the X axis at $A \left(h,0\right)$ and the Y axis at $B \left(0,k\right)$.

Then, X intercept $= OA = h$, Y intercept $= OB = k$

Given: $\dfrac{OA}{OB}= \dfrac{h}{k}=\dfrac{3}{10}$ $\implies$ $h = \dfrac{3 k}{10}$ $\;\;\; \cdots \; (1)$

Let the equation of the required line be: $\hspace{1em}$ $\dfrac{x}{h}+ \dfrac{y}{k} = 1$ $\;\;\; \cdots \; (2)$

The required line passes through $P \left(1,5\right)$

$\therefore$ $\;$ We have from equation $(2)$, $\hspace{1em}$ $\dfrac{1}{h} + \dfrac{5}{k} = 1$ $\;\;\; \cdots \; (3)$

Substituting the value of $h$ from equation $(1)$ in equation $(3)$ gives

$\dfrac{10}{3k} + \dfrac{5}{k} = 1$

i.e. $25 = 3k$ $\implies$ $k = \dfrac{25}{3}$ $\;\;\; \cdots \; (4a)$

Substituting the value of $k$ in equation $(1)$ gives

$h = \dfrac{3}{10} \times \dfrac{25}{3} = \dfrac{5}{2}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4a)$ and $(4b)$, equation $(2)$ becomes

$\dfrac{x}{5/2} + \dfrac{y}{25/3} = 1$

i.e. $\dfrac{2x}{5} + \dfrac{3y}{25} = 1$

i.e. $10 x + 3 y = 25$

$\therefore$ $\;$ The equation of the required line is: $\hspace{1em}$ $10x + 3 y = 25$

Locus

Find the points on the locus of points that are 3 units from X axis and 5 units from the point $(5,1)$.

Let $P \left(h,k\right)$ be a point on the required locus.

$P$ is 3 units from the X axis.

$\implies$ Equation of locus is: $\hspace{1em}$ $k = \pm 3$ $\;\;\; \cdots \; (1)$

Also, $P$ is 5 units from the point $(5,1)$.

i.e. $\sqrt{\left(h - 5\right)^2 + \left(k - 1\right)^2} = 5$

$\therefore$ $\;$ Equation of locus is: $\hspace{1em}$ $\left(h - 5\right)^2 + \left(k -1\right)^2 = 5^2$ $\;\;\; \cdots \; (2)$

Substituting $k = + 3$ in equation $(2)$ gives

$\left(h - 5\right)^2 + \left(3 - 1\right)^2 = 25$

i.e. $h^2 - 10 h + 25 + 4 = 25$

i.e. $h^2 - 10 h + 4 = 0$

i.e. $h = \dfrac{10 \pm \sqrt{100 - 16}}{2} = 5 \pm \sqrt{21}$

Substituting $k = - 3$ in equation $(2)$ gives

$h^2 - 10 h + 25 + \left(-3 - 1\right)^2 = 25$

i.e. $h^2 - 10 h + 16 = 0$

i.e. $\left(h - 2\right) \left(h - 8\right) = 0$

$\implies$ $h = 2$, $h = 8$

$\therefore$ $\;$ The points on the required locus are: $\hspace{1em}$ $\left(5 + \sqrt{21}, 3\right)$, $\left(5 - \sqrt{21}, 3\right)$, $\left(2, -3\right)$ and $\left(8, -3\right)$

Locus

If the points $P \left(6,2\right)$, $Q \left(-2,1\right)$ and $R$ are the vertices of $\triangle PQR$ and $R$ is the point on the locus $y = x^2 - 3x + 4$, then find the equation of locus of centroid of $\triangle PQR$.

Let $R \left(p,q\right)$ be the point on the locus.

$R$ satisfies the equation $y = x^2 - 3x + 4$

$\therefore$ $\;$ We have $\hspace{1em}$ $q = p^2 - 3p + 4$ $\;\;\; \cdots \; (1)$

Let $C \left(h,k\right)$ be a point on the equation of the locus of centroid of $\triangle PQR$.

Given: $P = \left(6,2\right)$ and $Q = \left(-2,1\right)$

$\therefore$ $\;$ Centroid of $\triangle PQR$ is

$\left(\dfrac{6 - 2 + p}{3}, \dfrac{2 + 1 + q}{3}\right) = \left(h, k\right)$

$\implies$ $\dfrac{4 + p}{3} = h$ $\implies$ $p = 3 h - 4$ $\;\;\; \cdots \; (2)$

and $\dfrac{3 + q}{3} = k$ $\implies$ $q = 3 k - 3$ $\;\;\; \cdots \; (3)$

Substituting the values of $p$ and $q$ from equations $(2)$ and $(3)$ in equation $(1)$ gives

$3 k - 3 = \left(3 h - 4\right)^2 - 3 \left(3 h - 4\right) + 4$

i.e. $3 k - 3 = 9 h^2 - 24 h + 16 - 9 h + 12 + 4$

i.e. $9 h^2 - 33 h - 3 k + 35 = 0$

$\therefore$ $\;$ The equation of locus of centroid of $\triangle PQR$ is: $\hspace{1em}$ $9x^2 - 33x - 3y + 35 = 0$

Locus

If $P \left(2,-7\right)$ is a given point and Q is a point on $2x^2 + 9y^2 = 18$, then find the equation of the locus of the midpoint of PQ.

Let $Q \left(a, b\right)$ be a point on $2x^2 + 9y^2 = 18$

Then, $2a^2 + 9b^2 = 18$ $\;\;\; \cdots \; (1)$

Coordinates of midpoint of $P \left(2,-7\right)$ and $Q \left(a,b\right)$ are $\left(\dfrac{2 + a}{2}, \dfrac{b - 7}{2}\right)$

Let $Z \left(h,k\right)$ be a point on the required locus.

Then, $h = \dfrac{a + 2}{2}$, $k = \dfrac{b - 7}{2}$

i.e. $a = 2 h - 2$ $\;\;\; \cdots \; (2a)$ and $b = 2 k + 7$ $\;\;\; \cdots \; (2b)$

Substituting the values of $a$ and $b$ from equations $(2a)$ and $(2b)$ in equation $(1)$ gives

$2 \left(2h - 2\right)^2 + 9 \left(2 k + 7\right)^2 = 18$

i.e. $2 \left(4 h^2 - 8h + 4\right) + 9 \left(4 k^2 + 28 k + 49\right) = 18$

i.e. $8 h^2 - 16 h + 8 + 36 k^2 + 252 k + 441 = 18$

i.e. $8 h^2 + 36 k^2 - 16 h + 252 k + 431 = 0$

$\therefore$ $\;$ The equation of the required locus is: $\hspace{1em}$ $8x^2 + 36y^2 - 16 x + 252 y + 431 = 0$

Locus

The coordinates of a moving point P are $\left(\dfrac{a}{2} \left(\text{cosec } \theta + \sin \theta\right), \dfrac{b}{2} \left(\text{cosec } \theta - \sin \theta\right)\right)$, where $\theta$ is a variable parameter. Find the equation of locus of P.

Let $P \left(h, k\right)$ be the moving point.

Then as per question,

$h = \dfrac{a}{2} \left(\text{cosec } \theta + \sin \theta\right)$, $k = \dfrac{b}{2} \left(\text{cosec } \theta - \sin \theta\right)$

i.e. $\dfrac{2h}{a} = \text{cosec } \theta + \sin \theta$ $\;\;\; \cdots (1)$

and $\dfrac{2k}{b} = \text{cosec } \theta - \sin \theta$ $\;\;\; \cdots \; (2)$

Squaring and subtracting equations $(1)$ and $(2)$ gives

$\dfrac{4h^2}{a^2} - \dfrac{4k^2}{b^2} = \left(\text{cosec } \theta + \sin \theta\right)^2 - \left(\text{cosec } \theta - \sin \theta\right)^2$

i.e. $4\left(\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2}\right) = \text{cosec}^2 \theta + \sin^2 \theta + 2 \text{cosec } \theta \sin \theta - \text{cosec}^2 \theta - \sin^2 \theta + 2 \text{cosec } \theta \sin \theta$

i.e. $4\left(\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2}\right) = 4 \text{cosec } \theta \sin \theta$

i.e. $\dfrac{h^2}{a^2} - \dfrac{k^2}{b^2} = 1$

i.e. $h^2 b^2 - k^2 a^2 = a^2 b^2$

$\therefore$ $\;$ The required equation of locus is: $\hspace{1em}$ $x^2 b^2 - y^2 a^2 = a^2 b^2$

Locus

Find the equation of the locus of point P such that the line segment AB, joining the points $A \left(1,-6\right)$ and $B \left(4,-2\right)$, subtends a right angle at P.

Let $P \left(h, k\right)$ be a point on the required locus.

Since the line segment AB subtends a right angle at P, we have

$PA^2 + PB^2 = AB^2$ $\;\;\; \cdots \; (1)$

Given: $A = \left(1,-6\right)$, $B = \left(4,-2\right)$

$\begin{aligned} \therefore \; PA^2 & = \left(h - 1\right)^2 + \left(k + 6\right)^2 \\\\ & = h^2 + k^2 - 2h + 12 k + 1 + 37 \\\\ & = h^2 + k^2 - 2h + 12 k + 37 \;\;\; \cdots \; (2a) \end{aligned}$

$\begin{aligned} PB^2 & = \left(h - 4\right)^2 + \left(k + 2\right)^2 \\\\ & = h^2 + k^2 - 8 h + 4 k + 16 + 4 \\\\ & = h^2 + k^2 - 8 h + 4k +20 \;\;\; \cdots \; (2b) \end{aligned}$

and, $AB^2 = \left(1-4\right)^2 + \left(-6 + 2\right)^2 = 25$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(2c)$. equation $(1)$ becomes

$h^2 + k^2 - 2 h + 12 k + 37 + h^2 + k^2 - 8 h + 4 k + 20 = 25$

i.e. $2h^2 + 2 k^2 - 10 h + 16 k + 32 = 0$

i.e. $h^2 + k^2 - 5 h + 8k + 16 = 0$

$\therefore$ $\;$ The required equation of locus is: $\hspace{1em}$ $x^2 + y^2 - 5 x + 8 y + 16 = 0$

Differential Equations

If the length of subnormal of a curve is constant and the curve passes through the origin, then find the equation of the curve.

Length of subnormal of a curve at any point $P \left(x,y\right)$ is $= y \; \dfrac{dy}{dx}$

Given: Length of subnormal $=$ constant

i.e. $y \; \dfrac{dy}{dx} = k$ $\;\;\;$ [k is a constant]

i.e. $y \; dy = k \; dx$

i.e. $\displaystyle \int y \; dy = k \int dx$

i.e. $\dfrac{y^2}{2} = k \; x + c$ $\;\;\; \cdots \; (1)$ $\;\;\;$ [c is the constant of integration]

Given: The required curve passes through the origin

$\therefore$ $\;$ Putting $x = 0$, $y = 0$ in equation $(1)$ gives

$c = 0$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{y^2}{2} = k \; x$ $\implies$ $y^2 = 2 \; k \; x$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the required equation of the curve.

Differential Equations

The increase in the principal amount kept at compound interest in a bank is proportional to the product of the principal amount and annual rate of interest.

1. Annual rate of interest in a bank is 5%. How many years will it take to double the principal amount?


2. At what annual rate of interest, the principal amount will double in 10 years?

Let the principal amount at any instant of time t be $= P$

Let the annual rate of interest $= R\% = \dfrac{R}{100} = r$

Increase in principal amount with time $= \dfrac{dP}{dt}$

$\therefore$ $\;$ As per question, $\;\;$ $\dfrac{dP}{dt} \propto Pr$

i.e. $\dfrac{dP}{dt} = k P r$ $\;\;\;$ [k is the constant of proportionality]

i.e. $\dfrac{dP}{P} = k \; r \; dt$

i.e. $\displaystyle \int \dfrac{dP}{p} = k \; r \int dt$

i.e. $\ln P = k \; r \; t + c$ $\;\;\;$ [c is the constant of integration] $\;\;\; \cdots \; (1)$

At time $t = 0$, initial amount $= P = P_0$

Substituting these values of $t$ and $P$ in equation $(1)$ give

$c = \ln P_0$ $\;\;\; \cdots \; (2)$

Substituting the value of $c$ from equation $(2)$ in equation $(1)$ gives

$\ln P = k \; r \; t + \ln P_0$

i.e. $\ln P - \ln P_0 = k \; r \; t$

i.e. $\ln \left(\dfrac{P}{P_0}\right) = k \; r \; t$

i.e. $\dfrac{P}{P_0} = e^{k\;r \; t}$ $\implies$ $P = P_0 \; e^{k \; r \; t}$ $\;\;\; \cdots \; (3)$

After 1 year, $t = 1$, $P = P_0 + P_0 \; r = P_0 \left(1 + r\right)$

Substituting these values of $t$ and $P$ in equation $(3)$ give

$P_0 \left(1 + r\right) = P_0 \; e^{k \; r}$

i.e. $e^{k \; r} = 1 + r$ $\implies$ $k \; r = \ln \left(1 + r\right)$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes

$P = P_0 \; e^{t \; \ln \left(1 + r\right)}$

i.e. $P = P_0 \; e^{\ln \left(1 + r\right)^t}$

i.e. $P = P_0 \left(1 + r\right)^t$ $\implies$ $P = P_0 \left(1 + \dfrac{R}{100}\right)^t$ $\;\;\; \cdots \; (5)$ $\;\;$ $\left[\because \; r = \dfrac{R}{100}\right]$

1. Given: Rate of interest $= R = 5\% = \dfrac{5}{100}$, $P = 2 P_0$
   
   Substituting these values in equation $(5)$ give
   
   $2 P_0 = P_0 \left(1 + \dfrac{5}{100}\right)^t$
   
   i.e. $2 = \left(1.05\right)^t$
   
   i.e. $\log \left(2\right) = t \; \log \left(1.05\right)$ $\implies$ $t = \dfrac{\log \left(2\right)}{\log \left(1.05\right)} = 14.2$
   
   $\therefore$ $\;$ It takes 14.2 years to double the principal when the rate of interest is 5%



2. Given: Time $= t = 10$ years, Principal amount $= P = 2 P_0$
   
   Substituting these values in equation $(5)$ give
   
   $2 P_0 = P_0 \left(1 + \dfrac{R}{100}\right)^{10}$
   
   i.e. $2 = \left(1 + \dfrac{R}{100}\right)^{10}$
   
   i.e. $\log \left(2\right) = 10 \; \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $\dfrac{1}{10} \times \log \left(2\right) = \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $\log \left(2^{0.1}\right) = \log \left(1 + \dfrac{R}{100}\right)$
   
   i.e. $2^{0.1} = 1 + \dfrac{R}{100}$ $\implies$ $R = 100 \times \left(2^{0.1} - 1\right) = 7.2$
   
   $\therefore$ $\;$ Annual rate of interest $= 7.2\%$

Differential Equations

In an experiment of culture of bacteria in a laboratory, the rate of increase of bacteria is proportional to the number of bacteria present at that time.

(i) If in one hour the number of bacteria gets doubled, then what is the number of bacteria at the end of 4 hours?

(ii) If the number of bacteria is 24,000 at the end of 3 hours, find the number of bacteria in the beginning.

Let the original number of bacteria at time $t = 0$ be $= N_0$

Let the number of bacteria at any instant of time $t$ be $= N$

Given: $\text{Rate of increase of bacteria} \propto \text{ Number present at that time}$

i.e. $\dfrac{dN}{dt} \propto N$

i.e. $\dfrac{dN}{dt} = k N$ $\hspace{2em}$ where $k$ is the constant of proportionality

i.e. $\dfrac{dN}{N} = k \; dt$

i.e. $\displaystyle \int \dfrac{dN}{N} = k \int dt$

i.e. $\log \left|N\right| = k \; t + c$ $\;\;\; \cdots \; (1)$ $\hspace{2em}$ where $c$ is the constant of integration

Applying the initial condition, when $t = 0$, $N = N_0$, equation $(1)$ becomes

$c = \log \left|N_0\right|$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$\log \left|N\right| = k \; t + \log \left|N_0\right|$

i.e. $\log \left|N\right| - \log \left|N_0\right| = k \; t$

i.e. $\log \left|\dfrac{N}{N_0}\right| = k \; t$

i.e. $\dfrac{N}{N_0} = e^{k t}$

i.e. $N = N_0 \; e^{kt}$ $\;\;\; \cdots \; (3)$

Now, in time $t = 1$ $\;$ hour, the number of bacteria gets doubled.

i.e. when $t = 1$, $N = 2 \; N_0$

$\therefore$ $\;$ In view of this condition equation $(3)$ becomes

$2 \; N_0 = N_0 \; e^k$

i.e. $e^{k} = 2$ $\implies$ $k = \log_e 2$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ becomes

$N = N_0 \; e^{t \; \log_e 2}$ $\;\;\; \cdots \; (5)$

Given: $t = 4$ hours

$\therefore$ $\;$ We have from equation $(5)$

$\begin{aligned} N & = N_0 \; e^{4 \; \log_e 2} \\\\ & = N_0 \; e^{\log_e \left(2^4\right)} \\\\ & = N_0 \; e^{\log_e 16} \\\\ & = 16 N_0 \end{aligned}$

i.e. Number of bacteria at the end of 4 hours $=$ 16 times the original number

Given: In time $t = 3$ hours, number of bacteria $= N = 24,000$

$\therefore$ $\;$ We have from equation $(5)$

$\begin{aligned} 24000 & = N_0 \; e^{\log_e 2} \\\\ & = N_0 \; e^{\log_e \left(2^3\right)} \\\\ & = N_0 \; e^{\log_e 8} \\\\ & = 8 N_0 \end{aligned}$

$\therefore$ $N_0 = \dfrac{24000}{8} = 3000$

$\therefore$ $\;$ Original number of bacteria $= 3000$ \end{enumerate}

Differential Equations

If the X intercept of the tangent to a curve at any point is four times its Y coordinate, then find the equation of the curve.

Let the slope of the tangent at any point $P \left(x,y\right)$ be $= \dfrac{dy}{dx}$

Let the equation of the tangent be $y = x \; \dfrac{dy}{dx} + k$ $\;\;\; \cdots \; (1)$ where k is the intercept.

Given: X intercept of tangent $= 4 y$

$y = 0$ gives the X intercept made by the tangent.

$\therefore$ We have

$0 = 4y \; \dfrac{dy}{dx} + k$

i.e. $k = - 4 \; y \; \dfrac{dy}{dx}$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$y = x \; \dfrac{dy}{dx} - 4 \; y \; \dfrac{dy}{dx}$

i.e. $y = \left(x - 4y\right) \dfrac{dy}{dx}$

i.e. $\dfrac{dx}{dy} = \dfrac{x - 4y}{y}$

i.e. $\dfrac{dx}{dy} = \dfrac{x}{y} - 4$ $\;\;\; \cdots \; (3)$

Let $\dfrac{x}{y} = u$ $\;\;\; \cdots \; (4a)$

$\implies$ $x = u \; y$ $\;\;\; \cdots \; (4b)$

Differentiating equation $(4b)$ w.r.t $y$ gives

$\dfrac{dx}{dy} = u + y \; \dfrac{du}{dy}$ $\;\;\; \cdots \; (4c)$

$\therefore$ $\;$ In view of equations $(4a)$ and $(4c)$, equation $(3)$ becomes

$u + y \; \dfrac{du}{dy} = u - 4$

i.e. $y \; \dfrac{du}{dy} = -4$

i.e. $\dfrac{du}{4} = - \dfrac{dy}{y}$ $\;\;\; \cdots \; (5)$

Integrating equation $(5)$ gives

$\dfrac{1}{4} \displaystyle \int du = - \int \dfrac{dy}{y}$

i.e. $\dfrac{u}{4} = - \log \left|y\right| + \log \left|c\right|$ $\;$ $c \;$ is the constant of integration

i.e. $\dfrac{u}{4} = \log \left|\dfrac{c}{y}\right|$

i.e. $\dfrac{x}{4y} = \log \left|\dfrac{c}{y}\right|$ $\;\;\;$ [from equation $(4a)$]

i.e. $e^{\frac{x}{4y}} = \dfrac{c}{y}$

i.e. $y = c \; e^{\frac{-x}{4y}}$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the required equation of the curve.

Differential Equations

Solve the differential equation $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$

The given differential equation is $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$

i.e. $y \; \dfrac{dx}{dy} - \left(x + 2y^2\right) = 0$

i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} - 2y = 0$

i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} = 2 y$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$

Here $P \left(y\right) = -\dfrac{1}{y}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = 2y$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$

i.e. $I.F = e^{\int -\frac{1}{y} \; dy} = e^{- \log \left|y\right|} = e^{\log \left|\frac{1}{y}\right|} = \dfrac{1}{y}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$\dfrac{x}{y} = \displaystyle \int 2 y \times \dfrac{1}{y} \; dy$

i.e. $\dfrac{x}{y} = 2 \displaystyle \int dy$

i.e. $\dfrac{x}{y} = 2y + c$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$

The given differential equation is $\;$ $\left(1 + y^2\right) \; dx = \left(\tan^{-1}\left(y\right) - x\right) \; dy$

i.e. $\dfrac{dx}{dy} = \dfrac{\tan^{-1}\left(y\right) - x}{1 + y^2}$

i.e. $\dfrac{dx}{dy} + \dfrac{x}{1 + y^2} = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$

Here $P \left(y\right) = \dfrac{1}{1 + y^2}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = \dfrac{\tan^{-1} \left(y\right)}{1 + y^2}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$

i.e. $I.F = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}\left(y\right)}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$x \; e^{\tan^{-1} \left(y\right)} = \displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy$ $\;\;\; \cdots \; (5)$

Let $\tan^{-1} \left(y\right) = p$ $\;\;\; \cdots \; (6a)$

Differentiating equation $(6a)$ gives

$\dfrac{dy}{1 + y^2} = dp$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(5)$ becomes

$\begin{aligned} \int \dfrac{\tan^{-1} \left(y\right)}{1 + y^2} \times e^{\tan^{-1}\left(y\right)} \; dy & = \int p \; e^{p} \; dp \\\\ & = p \int e^{p} \; dp - \int \left\{\int e^{p} \; dp \times \dfrac{d}{dp} \left(p\right) \right\} \; dp \\\\ & = p \; e^{p} - \int e^{p}\; dp \\\\ & = p \; e^{p} - e^{p} + c \\\\ & = e^{p} \left(p - 1\right) + c \\\\ & = e^{\tan{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c \;\;\; \cdots \; (7) \\ & \hspace{8em} \left[\text{In view of equation } (6a)\right] \end{aligned}$

$\therefore$ $\;$ We have from equations $(4)$ and $(7)$,

$x \; e^{\tan^{-1} \left(y\right)} = e^{\tan^{-1} \left(y\right)} \left(\tan^{-1} \left(y\right) - 1\right) + c$ $\;\;\; \cdots \; (8)$

Equation $(8)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $\dfrac{dy}{dx} + \dfrac{2y}{x} = e^x$

The given differential equation is $\;$ $\dfrac{dy}{dx} + \dfrac{2y}{x} = e^x$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P \left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = \dfrac{2}{x}$ $\;\;\; \cdots \; (2a)$; $\;\;$ $Q \left(x\right) = e^{x}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

$\begin{aligned} \text{i.e. I.F} = e^{\int \frac{2}{x} \; dx} & = e^{2 \log \left|x\right|} \\\\ & = e^{\log \left|x^2\right|} \\\\ & = x^2 \;\;\; \cdots \; (3) \end{aligned}$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$y \; x^2 = \displaystyle \int e^x \; x^2 \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now, } \int x^2 \; e^x \; dx & = x^2 \int e^x \; dx - \int \left\{\int e^x \; dx \times \dfrac{d}{dx} \left(x^2\right) \right\} \; dx \\\\ & = x^2 \; e^x - \int 2 \; x \; e^x \; dx \\\\ & = x^2 \; e^x - 2 \left[x \int e^x \; dx - \int \left\{\int e^x \; dx \times \dfrac{d}{dx} \left(x\right) \right\} \; dx\right] \\\\ & = x^2 \; e^x - 2 \left[x \; e^x - \int e^x \; dx\right] \\\\ & = x^2 \; e^x - 2 \; x \; e^x + 2 \; e^x + c \;\;\; \cdots \; (5) \end{aligned}$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$y \; x^2 = x^2 \; e^x - 2 \; x \; e^x + 2 \; e^x + c$

i.e. $y \; x^2 = e^x \left(x^2 - 2 \; x + 2\right) + c$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;$ $x \; \dfrac{dy}{dx} - y = \left(1 + x\right) e^{-x}$

The given differential equation is $\;$ $x \; \dfrac{dy}{dx} - y = \left(1 + x\right) e^{-x}$

i.e. $\dfrac{dy}{dx} - \dfrac{y}{x} = \dfrac{\left(1 + x\right) e^{-x}}{x}$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P \left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = - \dfrac{1}{x}$ $\;\;\; \cdots \; (2a)$; $\;\;$ $Q \left(x\right) = \dfrac{\left(1 + x\right) e^{-x}}{x}$ $\;\;\; \cdots \; (2b)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

$\begin{aligned} \text{i.e. I.F} = e^{\int \frac{-1}{x} \; dx} & = e^{- \log \left|x\right|} \\\\ & = e^{\log \left|\frac{1}{x}\right|} \\\\ & = \dfrac{1}{x} \;\;\; \cdots \; (3) \end{aligned}$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is

$\dfrac{y}{x} = \displaystyle \int \dfrac{\left(1 + x\right) e^{-x}}{x} \times \dfrac{1}{x} \; dx$

i.e. $\dfrac{y}{x} = \displaystyle \int \dfrac{\left(1 + x\right) e^{-x}}{x^2} \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now, } \int \dfrac{\left(1 + x\right) e^{-x}}{x^2} \; dx & = \left(1 + x\right) e^{-x} \int \dfrac{1}{x^2} \; dx \\ & \hspace{4em} - \int \left\{\int \dfrac{1}{x^2} \; dx \times \dfrac{d}{dx} \left[\left(1 + x\right) e^{-x}\right] \right\} \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} + \int \dfrac{1}{x} \left(- e^{-x} + e ^{-x} - x e^{-x}\right) \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} - \int e^{-x} \; dx \\\\ & = \dfrac{- \left(1 + x\right) e^{-x}}{x} + e^{-x} + c \;\;\; \cdots \; (5) \end{aligned}$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$\dfrac{y}{x} = \dfrac{- \left(1 + x\right) e^{-x}}{x} + e^{-x} + c$

i.e. $y = - \left(1 + x\right) e^{-x} + x e^{-x} + c x$

i.e. $y = - e^{-x} + cx$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;\;$ $\dfrac{dy}{dx} + 2\; y = \sin x$

The given differential equation is $\;\;$ $\dfrac{dy}{dx} + 2\; y = \sin x$ $\;\;\; \cdots \; (1)$

This is a linear differential equation in the form $\;$ $\dfrac{dy}{dx} + P\left(x\right)y = Q \left(x\right)$

Here $P \left(x\right) = 2$; $Q \left(x\right) = \sin x$ $\;\;\; \cdots \; (2)$

Integrating Factor (I.F) $= e^{\int P \left(x\right) \; dx}$

i.e. $\text{I.F} = e^{\int 2 \; dx} = e^{2x}$ $\;\;\; \cdots \; (3)$

According to the general solution,

$y \; e^{\int P \left(x\right) \; dx} = \displaystyle \int Q \left(x\right) \; e^{\int P \left(x\right) \; dx} \; dx$

$\therefore$ $\;$ In view of equations $(2)$ and $(3)$, the general solution of equation $(1)$ is

$y \; e^{2x} = \displaystyle \int \sin x \; e^{2x} \; dx$ $\;\;\; \cdots \; (4)$

$\begin{aligned} \text{Now } \int \sin x \; e^{2x} \; dx & = \sin x \int e^{2x} \; dx - \int \left\{\int e^{2x} \times \dfrac{d}{dx} \left(\sin x\right) \right\} \; dx \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \int e^{2x} \; \cos x \; dx \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \left[\cos x \int e^{2x} \; dx \right. \\ & \left. \hspace{4em} - \int \left\{\int e^{2x} \times \dfrac{d}{dx} \left(\cos x\right) \right\} \; dx\right] \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{1}{2} \left[\dfrac{e^{2x} \; \cos x}{2} + \dfrac{1}{2} \int e^{2x} \; \sin x \; dx\right] \\\\ & = \dfrac{e^{2x} \; \sin x}{2} - \dfrac{e^{2x} \; \cos x}{4} - \dfrac{1}{4} \int e^{2x} \; \sin x \; dx + c \end{aligned}$

i.e. $\left(1 + \dfrac{1}{4}\right) \displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{e^{2x}}{2} \left(\sin x - \dfrac{\cos x}{2}\right) + c$

i.e. $\displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{2}{5} \; e^{2x} \left(\sin x - \dfrac{\cos x}{2}\right) + c$

i.e. $\displaystyle \int \sin x \; e^{2x} \; dx = \dfrac{e^{2x} \left(2 \sin x - \cos x\right)}{5} + c$ $\;\;\; \cdots \; (5)$

$\therefore$ $\;$ In view of equation $(5)$, equation $(4)$ becomes

$y \; e^{2x} = \dfrac{e^{2x} \left(2 \sin x - \cos x\right)}{5} + c$

i.e. $y = \dfrac{1}{5} \left(2 \sin x - \cos x\right) + c e^{-2x}$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the general solution of the given differential equation.

Differential Equations

Find the particular solution of the differential equation $\;\;$ $\left(x^2 - 2y^2\right) \; dx + 2 \; x \; y \; dy = 0$ under the initial condition $y \left(1\right) = 1$

The given differential equation is $\;\;$ $\left(x^2 - 2y^2\right) \; dx + 2 \; x \; y \; dy = 0$

i.e. $2 \; x \; y \; dy = \left(2 y^2 - x^2\right) \; dx$

i.e. $\dfrac{dy}{dx} = \dfrac{2 y^2 - x^2}{2 \; x \; y}$

i.e. $\dfrac{dy}{dx} = \dfrac{y}{x} - \dfrac{x}{2y}$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} = v - \dfrac{1}{2 \; v}$

i.e. $x \; \dfrac{dv}{dx} = - \dfrac{1}{2 \; v}$

i.e. $2 \; v \; dv = - \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$2 \displaystyle \int v \; dv = - \int \dfrac{dx}{x}$

i.e. $2 \times \dfrac{v^2}{2} = - \log \left|x\right| + c$

i.e. $v^2 = - \log \left|x\right| + c$ $\;\;\; \cdots \; (4)$

Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives

$\left(\dfrac{y}{x}\right)^2 = - \log \left|x\right| + c$ $\;\;\; \cdots \; (5)$

Equation $(5)$ is the general solution of the given differential equation.

The initial condition is: when $x = 1$, $y = 1$

Substituting the initial condition in equation $(5)$ gives

$1 = - \log \left|1\right| + c$ $\implies$ $c = 1$

Substituting the value of $c$ in equation $(5)$ gives

$\left(\dfrac{y}{x}\right)^2 = - \log \left|x\right| + 1$

i.e. $\log \left|x\right| = 1 - \dfrac{y^2}{x^2}$

i.e. $x = e^{\left(1 - \frac{y^2}{x^2}\right)}$

i.e. $x = \dfrac{e}{e^{\left(\frac{y^2}{x^2}\right)}}$

i.e. $x \; e^{\left(\frac{y^2}{x^2}\right)} = e$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the particular solution of the given differential equation.

Differential Equations

Find the particular solution of the differential equation $\;\;$ $\left(x^2 + y^2\right) \; dx + x \; y \; dy = 0$ under the initial condition $y \left(1\right) = 1$

The given differential equation is $\;\;$ $\left(x^2 + y^2\right) \; dx + x \; y \; dy = 0$

i.e. $\dfrac{dy}{dx} = - \dfrac{\left(x^2 + y^2\right)}{x \; y}$

i.e. $\dfrac{dy}{dx} = - \left(\dfrac{x}{y} + \dfrac{y}{x}\right)$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} = - \dfrac{1}{v} - v$

i.e. $x \; \dfrac{dv}{dx} = - \left(\dfrac{1}{v} + 2v\right)$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{- \left(1 + 2 v^2\right)}{v}$

i.e. $\dfrac{v \; dv}{1 + 2 v^2} = - \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \dfrac{v}{1 + 2 v^2} \; dv = - \int \dfrac{dx}{x}$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{v \; dv}{1 + 2 v^2}$ $\;\;\; \cdots \; (5)$

Let $1 + 2 v^2 = p$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$4 \; v \; dv = dp$ $\implies$ $v \; dv = \dfrac{dp}{4}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5a)$ and $(5b)$, equation $(5)$ becomes

$\displaystyle \int \dfrac{v \; dv}{1 + 2 v^2} = \dfrac{1}{} \int \dfrac{dp}{p}$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ We have from equations $(4)$ and $(6)$

$\dfrac{1}{4} \displaystyle \int \dfrac{dp}{p} = - \int \dfrac{dx}{x}$

i.e. $\dfrac{1}{4} \; \log \left|p\right| = - \log \left|x\right| + \log \left|c_1\right|$ $\;\;\; \cdots \; (7)$

$\therefore$ $\;$ In view of equation $(5a)$, equation $(7)$ can be written as

$\dfrac{1}{4} \log \left|1 + 2v^2\right| + \log \left|x\right| = \log \left|c_1\right|$

i.e. $\log \left|x \left(1 + 2v^2\right)^{1/4}\right| = \log \left|c_1\right|$

i.e. $x \left(1 + 2 v^2\right)^{1/4} = c_1$ $\;\;\; \cdots \; (8)$

Substituting the value of $v$ from equation $(2a)$ in equation $(8)$ gives

$x \left(1 + \dfrac{2 \; y^2}{x^2}\right)^{1/4} = c_1$

i.e. $x^4 \left(1 + \dfrac{2 \; y^2}{x^2}\right) = \left(c_1\right)^4$

i.e. $x^4 \left(\dfrac{x^2 + 2\; y^2}{x^2}\right) = c$ $\;$ where $\left(c_1\right)^4 = c$

i.e. $x^2 \left(x^2 + 2 \; y^2\right) = c$ $\;\;\; \cdots \; (9)$

Equation $(9)$ is the general solution of the given differential equation.

The initial condition is: when $x = 1$, $y = 1$

Substituting the initial condition in equation $(9)$ gives

$1^2 \left(1^2 + 2 \times 1^2\right) = c$ $\implies$ $c = 3$

Substituting the value of $c$ in equation $(9)$ gives

$x^2 \left(x^2 + 2 \; y^2\right) = 3$ $\;\;\; \cdots \; (10)$

Equation $(10)$ is the particular solution of the given differential equation.

Differential Equations

Solve the differential equation $\;\;$ $\dfrac{dy}{dx} + \dfrac{y \left(x + y\right)}{x^2} = 0$

The given differential equation is $\;\;$ $\dfrac{dy}{dx} + \dfrac{y \left(x + y\right)}{x^2} = 0$

i.e. $\dfrac{dy}{dx} + \dfrac{y \; x + y^2}{x^2} = 0$

i.e. $\dfrac{dy}{dx} + \dfrac{y}{x} + \dfrac{y^2}{x^2} = 0$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} + v + v^2 = 0$

i.e. $x \; \dfrac{dv}{dx} = - 2 v - v^2$

i.e. $\dfrac{dv}{2 \; v + v^2} = - \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \dfrac{dv}{2 \; v + v^2} = - \int \dfrac{dx}{x}$

i.e. $\displaystyle \int \dfrac{dv}{v \left(2 + v\right)} = - \int \dfrac{dx}{x}$ $\;\;\; \cdots \; (4)$

Let $\dfrac{1}{v \left(2 + v\right)} = \dfrac{A}{v} + \dfrac{B}{2 + v}$ $\;\;\; \cdots \; (5)$

i.e. $1 = A \left(2 + v\right) + B \; v$ $\;\;\; \cdots \; (5a)$

In equation $(5a)$

$\begin{aligned} \text{When } & v = 0, & \implies 1 = 2 \; A & \implies A = \dfrac{1}{2} \;\;\; \cdots \; (5b) \\\\ \text{When } & v = -2, & \implies 1 = -2 \; B & \implies B = \dfrac{-1}{2} \;\;\; \cdots \; (5c) \end{aligned}$

$\therefore$ $\;$ In view of equations $(5b)$ and $(5c)$, equation $(5)$ becomes

$\dfrac{1}{v \left(2 + v\right)} = \dfrac{1 / 2}{v} - \dfrac{1 / 2}{2 + v}$ $\;\;\; \cdots \; (6)$

Integrating equation $(6)$ gives

$\displaystyle \int \dfrac{dv}{v \left(2 + v\right)} = \dfrac{1}{2} \int \dfrac{dv}{v} - \dfrac{1}{2} \int \dfrac{dv}{v + 2}$ $\;\;\; \cdots \; (6a)$

$\therefore$ $\;$ In view of equation $(6a)$, equation $(4)$ can be written as

$\displaystyle \dfrac{1}{2} \int \dfrac{dv}{v} - \dfrac{1}{2} \int \dfrac{dv}{v + 2} = - \int \dfrac{dx}{x}$

i.e. $\dfrac{1}{2} \log \left|v\right| - \dfrac{1}{2} \log \left|v + 2\right| = - \log \left|x\right| + \log \left|c_1\right|$

i.e. $\dfrac{1}{2} \log \left|\dfrac{v}{v + 2}\right| = \log \left|\dfrac{c_1}{x}\right|$

i.e. $\log \left|\dfrac{v}{v + 2}\right| = 2 \log \left|\dfrac{c_1}{x}\right|$

i.e. $\log \left|\dfrac{v}{v + 2}\right| = \log \left|\left(\dfrac{c_1}{x}\right)^2\right|$

i.e. $\dfrac{v}{v + 2} = \dfrac{c}{x^2}$ $\;\;\; \cdots \; (7)$ $\;$ where $\left(c_1\right)^2 = c$

Substituting the value of $v$ from equation $(2a)$ in equation $(7)$ gives

$\dfrac{\dfrac{y}{x}}{\dfrac{y}{x} + 2} = \dfrac{c}{x^2}$

i.e. $\dfrac{y}{y + 2x} = \dfrac{c}{x^2}$

i.e. $x^2 y = c \left(y + 2x\right)$ $\;\;\; \cdots \; (8)$

Equation $(8)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;\;$ $y \; dx + x \; \log \left(\dfrac{y}{x}\right) \; dy = 2 \; x \; dy$

The given differential equation is $\;\;$ $y \; dx + x \; \log \left(\dfrac{y}{x}\right) \; dy = 2 \; x \; dy$

i.e. $\dfrac{y}{x} \; dx + \log \left(\dfrac{y}{x}\right) \; dy = 2 \; dy$

i.e. $\left[2 - \log \left(\dfrac{y}{x}\right)\right] \; dy = \dfrac{y}{x} \; dx$

i.e. $\dfrac{dy}{dx} = \dfrac{\dfrac{y}{x}}{2 - \log \left(\dfrac{y}{x}\right)}$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

i.e. $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $x$ gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ can be written as

$v + x \; \dfrac{dv}{dx} = \dfrac{v}{2 - \log \left(v\right)}$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{v}{2 - \log \left(v\right)} - v$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{v - 2 \; v + v \; \log \left(v\right)}{2 - \log \left(v\right)}$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{v \; \log \left(v\right) - v}{2 - \log \left(v\right)}$

i.e. $\left[\dfrac{2 - \log \left(v\right)}{v \; \log \left(v\right) - v}\right] \; dv = \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \left[\dfrac{2 - \log \left(v\right)}{v \; \log \left(v\right) - v}\right] \; dv = \int \dfrac{dx}{x}$

i.e. $2 \displaystyle \int \dfrac{dv}{v \; \log \left(v\right) - v} - \int \dfrac{\log \left(v\right)}{v \; \log \left(v\right) - v} \; dv = \int \dfrac{dx}{x}$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{dv}{v \; \log \left(v\right) - v} = \int \dfrac{dv}{v \left[\log \left(v\right) - 1\right]}$ $\;\;\; \cdots \; (5)$

Let $\log \left(v\right) - 1 = p$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$\dfrac{dv}{v} = dp$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5a)$ and $(5b)$, equation $(5)$ becomes

$\displaystyle \int \dfrac{dv}{v \; \log \left(v\right) - v} = \int \dfrac{dp}{p}$ $\;\;\; \cdots \; (5c)$

Consider $\displaystyle \int \dfrac{\log \left(v\right)}{v \; \log \left(v\right) - v} \; dv$ $\;\;\; \cdots \; (6)$

Let $v \; \log \left(v\right) - v = u$ $\;\;\; \cdots \; (6a)$

Differentiating equation $(6a)$ gives

$\left[v \times \dfrac{1}{v} + \log \left(v\right) - 1\right] \; dv = du$

i.e. $\log \left(v\right) \; dv = du$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(6)$ becomes

$\displaystyle \int \dfrac{\log \left(v\right)}{v \; \log \left(v\right) - v} \; dv = \int \dfrac{du}{u}$ $\;\;\; \cdots \; (6c)$

$\therefore$ $\;$ In view of equations $(5c)$ and $(6c)$, equation $(4)$ becomes

$2 \displaystyle \int \dfrac{dp}{p} - \int \dfrac{du}{u} = \int \dfrac{dx}{x}$

i.e. $2 \log \left|p\right| - \log \left|u\right| = \log \left|x\right| + \log \left|c\right|$

i.e. $\log \left|\dfrac{p^2}{u}\right| = \log \left|c \; x\right|$

i.e. $\dfrac{p^2}{u} = c \; x$ $\;\;\; \cdots \; (7)$

Substituting the values of $p$ and $u$ from equations $(5a)$ and $(6a)$, respectively, in equation $(7)$, we have

$\dfrac{\left[\log \left|v\right| - 1\right]^2}{v \; \log \left|v\right| - v} = c \; x$

i.e. $\dfrac{\left[\log \left|v\right| - 1\right]^2}{v \left[\log \left|v\right| - 1\right]} = c \; x$

i.e. $\dfrac{\log \left|v\right| - 1}{v} = c \; x$

i.e. $\log \left|v\right| - 1 = c \; x \; v$ $\;\;\; \cdots \; (8)$

In view of equation $(2a)$, equation $(8)$ becomes

$\log \left|\dfrac{y}{x}\right| - 1 = c \times x \times \dfrac{y}{x}$

i.e. $\log \left|\dfrac{y}{x}\right| - 1 = c \; y$ $\;\;\; \cdots \; (9)$

Equation $(9)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\;\;$ $y + 2 \; y \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; x \; e^{x/y}$

The given differential equation is $\;\;$ $y + 2 \; y \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; x \; e^{x/y}$

i.e. $1 + 2 \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; \dfrac{x}{y} \; e^{x/y}$

i.e. $2 \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; \dfrac{x}{y} \; e^{x/y} - 1$ $\;\;\; \cdots \; (1)$

Let $\dfrac{x}{y} = v$ $\;\;\; \cdots \; (2a)$

i.e. $x = v \; y$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t $y$ gives

$\dfrac{dx}{dy} = v + y \; \dfrac{dv}{dy}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes

$2 \; e^v \left(v + y \; \dfrac{dv}{dy}\right) = 2 \; v \; e^v - 1$

i.e. $v + y \; \dfrac{dv}{dy} = \dfrac{2 \; v \; e^v - 1}{2 \; e^v}$

i.e. $v + y \; \dfrac{dv}{dy} = v - \dfrac{1}{2 \; e^v}$

i.e. $y \; \dfrac{dv}{dy} = \dfrac{-1}{2 \; e^v}$

i.e. $2 \; e^v \; dv = - \dfrac{dy}{y}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$2 \displaystyle \int e^v \; dv = - \int \dfrac{dy}{y}$

i.e. $2 \; e^v = - \log \left|y\right| + \log \left|c\right|$ $\;\;\; \cdots \; (4)$

Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives

$2 \; e^{x/y} = \log \left|\dfrac{c}{y}\right|$ $\;\;\; \cdots \; (5)$

Equation $(5)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation
$\;$ $\left[x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)\right] y = \left[y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)\right] x \; \dfrac{dy}{dx}$

The given differential equation is

$\left[x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)\right] y = \left[y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)\right] x \; \dfrac{dy}{dx}$

i.e. $\dfrac{dy}{dx} = \left[\dfrac{x \cos \left(\dfrac{y}{x}\right) + y \sin \left(\dfrac{y}{x}\right)}{y \sin \left(\dfrac{y}{x}\right) - x \cos \left(\dfrac{y}{x}\right)}\right] \times \dfrac{y}{x}$

i.e. $\dfrac{dy}{dx} = \left[\dfrac{\cos \left(\dfrac{y}{x}\right) + \dfrac{y}{x} \sin \left(\dfrac{y}{x}\right)}{\dfrac{y}{x} \sin \left(\dfrac{y}{x}\right) - \cos \left(\dfrac{y}{x}\right)}\right] \times \dfrac{y}{x}$ $\;\;\; \cdots \; (1)$

Let $\dfrac{y}{x} = v$ $\;\;\; \cdots \; (2a)$

$\implies$ $y = v \; x$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2b)$ w.r.t x gives

$\dfrac{dy}{dx} = v + x \; \dfrac{dv}{dx}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes

$v + x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}\right] \times v$

i.e. $x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)} - 1\right] \times v$

i.e. $x \; \dfrac{dv}{dx} = \left[\dfrac{\cos \left(v\right) + v \; \sin \left(v\right) - v \; \sin \left(v\right) + \cos \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}\right] \times v$

i.e. $x \; \dfrac{dv}{dx} = \dfrac{2 \; v \; \cos \left(v\right)}{v \; \sin \left(v\right) - \cos \left(v\right)}$

i.e. $\left[\dfrac{v \; \sin \left(v\right) - \cos \left(v\right)}{v \; \cos \left(v\right)}\right] \; dv = 2 \; \dfrac{dx}{x}$

i.e. $\left[\tan \left(v\right) - \dfrac{1}{v}\right] \; dv = 2 \; \dfrac{dx}{x}$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \left[\tan \left(v\right) - \dfrac{1}{v}\right] \; dv = 2 \int \dfrac{dx}{x}$

i.e. $\log \left|\sec \left(v\right)\right| - \log \left|v\right| = 2 \; \log \left|x\right| + \log \left|c\right|$ $\;\;\; \cdots \; (4)$

Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives

$\log \left|\sec \left(\dfrac{y}{x}\right)\right| - \log \left|\dfrac{y}{x}\right| - 2 \log \left|x\right| = \log \left|c\right|$

i.e. $\log \left|\dfrac{\sec \left(\dfrac{y}{x}\right)}{\dfrac{y}{x} \times x^2}\right| = \log \left|c\right|$

i.e. $\dfrac{\sec \left(\dfrac{y}{x}\right)}{x \; y} = c$

i.e. $\sec \left(\dfrac{y}{x}\right) = c \; x \; y$ $\;\;\; \cdots \; (5)$

Equation $(5)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\left(x + y\right)^2 \; \dfrac{dy}{dx} = a^2$

The given differential equation is $\;\;$ $\left(x + y\right)^2 \; \dfrac{dy}{dx} = a^2$ $\;\;\; \cdots \; (1)$

Let $x + y =u$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2a)$ w.r.t x gives

$1 + \dfrac{dy}{dx} = \dfrac{du}{dx}$

i.e. $\dfrac{dy}{dx} = \dfrac{du}{dx} - 1$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$u^2 \left(\dfrac{du}{dx} -1\right) = a^2$

i.e. $\dfrac{du}{dx} - 1 = \dfrac{a^2}{u^2}$

i.e. $\dfrac{du}{dx} = \dfrac{a^2}{u^2} + 1$

i.e. $\dfrac{du}{dx} = \dfrac{a^2 + u^2}{u^2}$

i.e. $\left(\dfrac{u^2}{u^2 + a^2}\right) du = dx$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \dfrac{u^2}{u^2 + a^2} \; du = \int dx$

i.e. $\displaystyle \int \dfrac{u^2 + a^2}{u^2 + a^2} \; du - \int \dfrac{a^2}{u^2 + a^2} \; du = \int dx$

i.e. $\displaystyle \int du - a^2 \int \dfrac{du}{u^2 + a^2} = \int dx$

i.e. $u - a^2 \times \dfrac{1}{a} \tan^{-1}\left(\dfrac{u}{a}\right) = x + c$ $\;\;\; \cdots \; (4)$

Substituting the value of $u$ from equation $(2a)$ gives

i.e. $x + y - a \tan^{-1} \left(\dfrac{x + y}{a}\right) = x + c$

i.e. $y - a \tan^{-1} \left(\dfrac{x + y}{a}\right) = c$ $\;\;\; \cdots \; (5)$

Equation $(5)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $\dfrac{dy}{dx} = e^{x + y}$

The given differential equation is $\;$ $\dfrac{dy}{dx} = e^{x + y}$ $\;\;\; \cdots \; (1)$

Let $x + y = u$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2a)$ gives

$1 + \dfrac{dy}{dx} = \dfrac{du}{dx}$

i.e. $\dfrac{dy}{dx} = \dfrac{du}{dx} -1$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\dfrac{du}{dx} - 1 = e^u$

i.e. $\dfrac{du}{dx} = e^u + 1$

i.e. $\dfrac{du}{e^u + 1} = dx$ $\;\;\; \cdots \; (3)$

Integrating equation $(3)$ gives

$\displaystyle \int \dfrac{du}{e^u + 1} = \int dx$ $\;\;\; \cdots \; (4)$

Consider $\displaystyle \int \dfrac{du}{e^u + 1}$ $\;\;\; \cdots \; (5)$

Let $e^u + 1 = t$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$e^u \; du = dt$

i.e. $du = \dfrac{dt}{e^u} = \dfrac{dt}{t - 1}$ $\;\;\; \cdots \; (5b)$ $\;\;$ [by equation $(5a)$]

$\therefore$ $\;$ In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes

$\displaystyle \int \dfrac{dt / t - 1}{t} = \int dx$

i.e. $\displaystyle \int \dfrac{dt}{t \left(t - 1\right)} = \int dx$ $\;\;\; \cdots \; (6)$

Let $\dfrac{1}{t \left(t - 1\right)} = \dfrac{A}{t} + \dfrac{B}{t - 1}$ $\;\;\; \cdots \; (7)$

i.e. $1 = A \left(t - 1\right) + B \; t$ $\;\;\; \cdots \; (7a)$

Putting $t = 0$ in equation $(7a)$ gives

$1 = - A$ $\implies$ $A = -1$ $\;\;\; \cdots \; (7b)$

Putting $t = 1$ in equation $(7a)$ gives

$B = 1$ $\;\;\; \cdots \; (7c)$

$\therefore$ $\;$ In view of equations $(7b)$ and $(7c)$, equation $(7)$ becomes

$\dfrac{1}{t \left(t - 1\right)} = \dfrac{-1}{t} + \dfrac{1}{t - 1}$ $\;\;\; \cdots \; (8)$

Integrating equation $(8)$ gives

$\displaystyle \int \dfrac{dt}{t \left(t - 1\right)} = \int \dfrac{-dt}{t} + \int \dfrac{dt}{t - 1}$ $\;\;\; \cdots \; (8a)$

$\therefore$ $\;$ In view of equation $(8a)$, equation $(6)$ can be written as

$\displaystyle \int \dfrac{-dt}{t} + \int \dfrac{dt}{t - 1} = \int dx$

i.e. $- \log \left|t\right| + \log \left|t - 1\right| + \log \left|c\right| = x$

i.e. $\log \left|\dfrac{c \left(t - 1\right)}{t}\right| = x$

i.e. $\dfrac{c \left(t - 1\right)}{t} = e^x$ $\;\;\; \cdots \; (9)$

Substituting the value of $t$ from equation $(5a)$ in equation $(9)$ gives

$\dfrac{c \left(e^u + 1 - 1\right)}{e^u + 1} = e^x$

i.e. $\dfrac{c \; e^u}{e^u + 1} = e^x$ $\;\;\; \cdots \; (10)$

In view of equation $(2a)$, equation $(10)$ becomes

$\dfrac{c \; e^{x+y}}{e^{x+y} + 1} = e^x$

i.e. $\dfrac{c \; e^y}{e^{x+y} + 1} = 1$

i.e. $c \; e^y = e^{x+y} + 1$ $\;\;\; \cdots \; (11)$

Equation $(11)$ is the general solution of the given differential equation.

Differential Equations

Solve the differential equation $e^{dy/dx} = x + 1$.

Also find the particular solution when the initial conditions are $y \left(0\right) = 3$, $x > -1$.

Given: $e^{dy/dx} = x + 1$

i.e. $\log \left|e^{dy/dx}\right| = \log \left|x + 1\right|$

i.e. $\dfrac{dy}{dx} = \log \left|x + 1\right|$

i.e. $dy = \log \left|x + 1\right| \; dx$ $\;\;\; \cdots \; (1)$

Integrating equation $(1)$ gives

$\displaystyle \int dy = \int \log \left|x + 1\right| \; dx$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{Now, } \int \log \left|x + 1\right| \; dx & = \log \left|x + 1\right| \int dx - \int \left\{\int dx \times \dfrac{d}{dx} \left(\log \left|x + 1\right|\right) \right\} \; dx \\\\ & = x \log \left|x + 1\right| - \int x \times \dfrac{1}{x +1} \; dx \\\\ & = x \log \left|x + 1\right| - \left[\int \dfrac{x + 1}{x + 1} \; dx - \int \dfrac{dx}{x + 1}\right] \\\\ & = x \log \left|x + 1\right| - \int dx + \int \dfrac{dx}{x + 1} \\\\ & = x \log \left|x + 1\right| - x + \log \left|x + 1\right| + c_1 \\\\ & = \left(x + 1\right) \log \left|x + 1\right| - x + c_1 \;\;\; \cdots \; (3) \end{aligned}$

and $\displaystyle \int dy = y + c_2$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equations $(3)$ and $(4)$, equation $(2)$ becomes

$y + c_2 = \left(x + 1\right) \log \left|x + 1\right| - x + c_1$

i.e. $y = \left(x + 1\right) \log \left|x + 1\right| - x + c$ $\;\;\; \cdots \; (5)$ $\;$ where $c = c_1 - c_2$

Equation $(5)$ is the general solution of the given differential equation.

The initial condition is $y \left(0\right) = 3$

i.e. when $x = 0$, $y = 3$

Substituting the values of x and y in equation $(5)$ gives

$3 = \left(0 + 1\right) \log \left|0 + 1\right| - 0 + c$

i.e. $3 = \log \left|1\right| + c$ $\implies$ $c = 3$ $\;\;\; \cdots \; (6)$

Substituting the value of c from equation $(6)$ in equation $(5)$ gives

$y = \left(x + 1\right) \log \left|x + 1\right| - x + 3$ $\;\;\; \cdots \; (7)$

Equation $(7)$ is the particular solution of the given differential equation when $y \left(0\right) = 3$.

Differential Equations

Solve the differential equation $x \; y \; \dfrac{dy}{dx} = y + 2$.
Also find the particular solution when the initial conditions are $y \left(2\right) = 0$.

The given differential equation is $\;\;$ $x \; y \; \dfrac{dy}{dx} = y + 2$

i.e. $\dfrac{y \; dy}{y + 2} = \dfrac{dx}{x}$ $\;\;\; \cdots \; (1)$

Integrating equation $(1)$ gives

$\displaystyle \int \dfrac{y \; dy}{y + 2} = \int \dfrac{dx}{x}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{Now, } \int \dfrac{y}{y + 2} dy & = \int \dfrac{y + 2}{y + 2} \; dy - 2 \int \dfrac{dy}{y + 2} \\\\ & = \int dy - 2 \int \dfrac{dy}{y + 2} \\\\ & = y - 2 \log \left|y + 2\right| + \log \left|c_1\right| \;\;\; \cdots \; (3a) \end{aligned}$

and $\displaystyle \int \dfrac{dx}{x} = \log \left|x\right| + \log \left|c_2\right|$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ In view of equations $(3a)$ and $(3b)$, equation $(2)$ becomes

$y - 2 \log \left|y + 2\right| + \log \left|c_1\right| = \log \left|x\right| + \log \left|c_2\right|$

i.e. $y - 2 \log \left|y + 2\right| = \log \left|x\right| + \log \left|c_2\right| - \log \left|c_1\right|$

i.e. $y - 2 \log \left|y + 2\right| = \log \left|x\right| + \log \left|\dfrac{c_2}{c_1}\right|$

i.e. $y - 2 \log \left|y + 2\right| = \log \left|x\right| + \log \left|c\right|$ $\;$ where $c = \dfrac{c_2}{c_1}$

i.e. $y = \log \left|x\right| + \log \left|\left(y + 2\right)^2\right| + \log \left|c\right|$

i.e. $y = \log \left|c \; x \; \left(y + 2\right)^2\right|$

i.e. $e^y = c \; x \; \left(y + 2\right)^2$ $\;\;\; \cdots \; (4)$

Equation $(4)$ is the general solution of the given differential equation.

Given: when $x = 2$, $y = 0$

Substituting the values of x and y in equation $(4)$ gives

$e^0 = c \times 2 \times \left(0 + 2\right)^2$

i.e. $1 = 8 c$ $\implies$ $c = \dfrac{1}{8}$ $\;\;\; \cdots \; (5)$

Substituting the value of c from equation $(5)$ in equation $(4)$ gives

$e^y = \dfrac{1}{8} \times x \times \left(y + 2\right)^2$

i.e. $8 e^y = x \left(y + 2\right)^2$ $\;\;\; \cdots \; (6)$

Equation $(6)$ is the particular solution of the given differential equation when $y \left(2\right) = 0$.