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Find the particular solution of the differential equation \frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}.  given that y= 1  when x= 0.

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

As the Differential equation can be written as
\frac{dy}{dx}= \frac{xy}{x^{2}+y^{2}}= \frac{\frac{y}{x}}{1+\frac{y^{2}}{x^{2}}}    so its homogeneous
put  y= v{x}\Rightarrow \frac{dy}{dx}= v+\frac{xdv}{dx}
     \therefore v+x\frac{dv}{dx}= \frac{v}{1+v^{2}}
\Rightarrow x\, \frac{dv}{dx}= \frac{v}{1+v^{2}}-v= \frac{v-v-v^{3}}{1+v^{2}}
\Rightarrow \int \frac{1+v^{2}}{v^{3}}dv= -\int \frac{dx}{x}
\Rightarrow \int\left ( \frac{1}{v^{3}} +\frac{1}{v}\right )dv= -\int \frac{dx}{x}
\Rightarrow -\frac{1}{2v^{2}}+\log \left | v \right |= -\log \left | x \right |+c
\Rightarrow -\frac{x^{2}}{2y^{2}}+\log \left | \frac{y}{x} \right |= -\log \left | x \right |+c
\Rightarrow -\frac{x^{2}}{2y^{2}}+\log \left | y \right |= c
As y= 1  when x= 0,  so  \frac{-0^{2}}{2\times 12}+\log \left | 1 \right |= c\Rightarrow c= 0
Here, the required solution is \frac{-x^{2}}{2y^{2}}+\log \left | y \right |= 0
or   \log \left | y \right |= \frac{x^{2}}{2y^{2}}

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