Question

The general solution of the differential equation $\frac{dy}{dx}=\frac{r^2}{{(x+y)}^2},$ is (where $r$ is a fixed constant and $c$ is a constant of integration )

• A. $(x^2+y^2)-r{\tan }^{-1}\left(\frac{x-y}{r}\right)=y+c$
• B. $x-r{\tan }^{-1}\left(\frac{x+y}{r}\right)=c$
• C. $(x^2+y^2)-r{\tan }^{-1}\left(\frac{x-y}{r}\right)=x+c$
• D. $y-r{\tan }^{-1}\left(\frac{x+y}{r}\right)=c$

Answer 1

The correct option is D $y-r{\tan }^{-1}\left(\frac{x+y}{r}\right)=c$

Substituting $x+y=v,$ we have
$\frac{dy}{dx}=\frac{dv}{dx}-1$
and thus the equation reduces to
$\frac{dv}{dx}-1=\frac{r^2}{v^2}$
$\Rightarrow \frac{v^2}{r^2+v^2}dv=dx$
$\Rightarrow (1-\frac{r^2}{r^2+v^2})dv=dx$
Integrating , we have
$\Rightarrow v-r{\tan }^{-1}\left(\frac{v}{r}\right)=x+c$
$\Rightarrow (x+y)-r{\tan }^{-1}\left(\frac{x+y}{r}\right)=x+c$
$\Rightarrow y-r{\tan }^{-1}\left(\frac{x+y}{r}\right)=c$