Solve the differential equation: <div style="float:left; wid

Solve the differential equation:
$\frac{dy}{dx}= \frac{x+y}{x-y}$

Answers (1)

$\frac{dy}{dx}= \frac{x+y}{x-y}\: \: \left ( given \right )$
$y= vx$
$v+\frac{xdv}{dx}= \frac{x+vx}{x-vx}$ $v+\frac{xdv}{dx}= v+\frac{xdv}{dx}$
$v+\frac{xdv}{dx}= \frac{\not{x}\left ( 1+v \right )}{\not{x}\left ( 1-v \right )}\: \: \Rightarrow \: \frac{xdv}{dx}= \frac{1+v}{1-v}-v$
Taking LCM
$\frac{xdv}{dx}= \frac{1+v-v\left ( 1-v \right )}{\left ( 1-v \right )}\: \:\Rightarrow \frac{1+v-v+v^{2}}{\left ( 1-v \right )}$
$\frac{xdv}{dx}\Rightarrow \frac{1+v^{2}}{\left ( 1-v \right )}$
$\frac{1-v}{1+v^{2}}dv= \frac{1}{x}dx$
Integrating Both sides.
$\Rightarrow \int \frac{1-v}{1+v^{2}}dv\int \frac{1}{x}dx$
$\Rightarrow \int \frac{1}{1+v^{2}}dv-\frac{1}{2}\int \frac{2v}{1+v^{2}}dv= \log x+c$

(LHS) = (RHS)---------(1)
$\Rightarrow \tan^{-1}v-\frac{1}{2}\log \left | 1+v^{2} \right |\; \; \; \; \left [ assuming \: 1+v^{2}= t \: \: 2vdv= dt \right ]$
$\therefore \int \frac{1-v}{1+v^{2}}dx\: \Rightarrow \: \tan^{-1}v-\frac{1}{2}\log \left | 1+v^{2} \right |----(2)$
Substituing (2) in (1) RHS
$\tan^{-1}v-\frac{1}{2}\log \left | 1+v^{2} \right |= \log x+c$
substituing $v= \frac{y}{x}$
$\tan^{-1}\frac{y}{x}-\frac{1}{2}\log \left | \frac{x^{2}+y^{2}}{x^{2}} \right |= \log x+c$