Write the integrating factor of the differential equation \left ( \tan^{-1}y-x \right )dy= \left ( 1+y^{2} \right )dx.

 

 

 

 
 
 
 
 

Answers (1)

Re-writing the D.E: \frac{dx}{dy}+\frac{1}{\left ( 1+y^{2} \right )}\cdot x= \frac{\tan^{-1}y}{1+y^{2}}
which is of the form \frac{dy}{dx}+p\left ( y \right )x= Q\left ( y \right )
\therefore p= \frac{1}{1+y^{2}}\; \S \; Q= \frac{\tan^{-1}y}{1+y^{2}}
Therefore the I.F.  = e^{\int \frac{1}{1+y^{2}}\: dy}= e^{\tan^{-1}y}

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