Question

The solution of the differential equation $\frac{dy}{dx}=\frac{x^2+xy+y^2}{x^2}$, is
(a) ${\tan }^{-1}\left(\frac{x}{y}\right)=\log y+C$

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

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

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

Answer 1

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

We have,
$\frac{dy}{dx}=\frac{x^2+xy+y^2}{x^2}\left(1\right)$
This is homogenous differential equation.
$$\begin{gathered} \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Now},\mathrm{putting}\frac{dy}{dx}=v+x\frac{dv}{dx}\mathrm{and}y=vx\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{x^2+x^{2}v+x^2{v}^2}{x^2} \\ \Rightarrow v+x\frac{dv}{dx}=1+v+v^2 \\ \Rightarrow x\frac{dv}{dx}=1+v^2 \\ \Rightarrow \left(\frac{1}{1+v^2}\right)dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \int \frac{1}{1+v^2}dv=\int \frac{1}{x}dx \\ \Rightarrow {\tan }^{-1}v=\log x+C \\ \Rightarrow {\tan }^{-1}\left(\frac{y}{x}\right)=\log x+C \end{gathered}$$