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  • Find the particular solution of the differential equation<img alt="\left ( 1+x^{2} \right )\frac{dy}{dx}+2xy= \frac{1}{1+x^{2}}," src="https://entrancecorner.codecogs.com/gif.latex?%5Cleft%20%28%201&plus;x%5E%7B2%7D%20%5Cright%20%29%5Cfrac%7Bdy%7D%

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

 

 

 

 
 
 
 
 

Answers (1)

\left ( 1+x^{2} \right )\frac{dy}{dx}+2xy= \frac{1}{1+x^{2}}\Rightarrow \frac{dy}{dx}+\left ( \frac{2x}{1+x^{2}} \right )y= \frac{1}{\left ( 1+x^{2} \right )^{2}}
On comparing with \frac{dy}{dx}+p\left ( x \right )y= Q\left ( x \right ) we get
p\left ( x \right )= \frac{2x}{1+x^{2}},Q\left ( x \right )= \frac{1}{\left ( 1+x^{2} \right )^{2}}
Now IF= e^{\int \frac{2x}{1+x^{2}}dx}= e^{\log \left ( 1+x^{2} \right )}= 1+x^{2}
so,the solution is given by :\left ( 1+x^{2} \right )y= \int (1+x^{2})\times \frac{1}{\left ( 1+x^{2} \right )^2}dx+c
\Rightarrow \left ( 1+x^{2} \right )y= \int \frac{1}{1+x^{2}}dx+c= \tan^{-1}x+c
Now, it is given that y = 0, when x =1, so \left ( 1+1^{2} \right ) \cdot 0= \tan^{-1}1+c
0= \tan^{-1}1+c\Rightarrow c= \frac{-\pi }{4}
Hence the reuired particular solution is
\left ( 1+x^{2} \right )y= \tan^{-1}x - \frac{\pi }{4}

Posted by

Ravindra Pindel

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