Question

Find particular solution of differential equation $x^{2}dy-(3x^2+xy+y^2)dx=0,y\left(1\right)=1$

Answer 1

$x^{2}dy-(3x^2+xy+y^2)dx=0$

$\Rightarrow x^{2}dy=(3x^2+xy+y^2)dx$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2+xy+y^2}{x^2}$

$\Rightarrow \frac{dy}{dx}=3+\frac{xy+y^2}{x^2}$

Let $y=vx$

$\Rightarrow \frac{dy}{dx}=x\frac{dv}{dx}+v$

Therefore,

$x\frac{dv}{dx}+v=3+\frac{v{x}^2+v^2{x}^2}{x^2}$

$x\frac{dv}{dx}+v=3+v+v^2$

$\Rightarrow \frac{dv}{3+v^2}=\frac{dx}{x}$

Integrating bothe sides, we have

$\Rightarrow \int \frac{dv}{3+v^2}=\int \frac{dx}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{v}{\sqrt{3}}\right)=\log x+C$

Substituting the value of $v$ in above equation, we get

$\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{y}{\sqrt{3}x}=\log x+C.....\left(1\right)$

$\because y\left(1\right)=1\left(\text{Given}\right)$

$\therefore \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\log 1+C$

$\Rightarrow C=\frac{\sqrt{3}\pi }{18}$

Substituting the value of $C$ in equation $\left(1\right)$, we get

$\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{y}{\sqrt{3}x}=\log x+\frac{\sqrt{3}\pi }{18}$

Hence the particular solution for the given differential equation is-

$\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{y}{\sqrt{3}x}=\log x+\frac{\sqrt{3}\pi }{18}$