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

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