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.