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.