Question

The solution of the differential equation $x^2\frac{dy}{dx}=x^2+xy+y^2$ is-

• A. $ta{n}^{-1}\frac{y}{x}=logx+c$
• B. $ta{n}^{-1}\frac{y}{x}=-logx+c$
• C. $si{n}^{-1}\frac{y}{x}=logx+c$
• D. $ta{n}^{-1}\frac{x}{y}=logx+c$

Answer 1

The correct option is A $ta{n}^{-1}\frac{y}{x}=logx+c$

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

$\frac{dy}{dx}=1+\frac{y}{x}+(\frac{y}{x}{)}^2$

Let $\frac{y}{x}=v$

$y=vx$

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

$v+x\frac{dv}{dx}=1+v+v^2$

$x\frac{dv}{dx}=1+v^2$

$\frac{dv}{1+v^2}=\frac{dx}{x}$

${\tan }^{-1}v=\log x+C$

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