Find the particular solution of the differential equation :
\left ( 1+e^{2x} \right )dy+\left ( 1+y^{2} \right )e^{x}dx= 0,  given that y\left ( 0 \right )= 1\cdot

 

 

 

 
 
 
 
 

Answers (1)

\left ( 1+e^{2x} \right )dy+\left ( 1+y^{2} \right )e^{x}dx= 0
\Rightarrow \int \frac{dy}{1+y^{2}}+\int \frac{e^{x}dx}{1+e^{2x}}= 0
put e^{x}= t\; \; \Rightarrow e^{x}dx= dt  in the second integral
\therefore \tan^{-1}y+\int \frac{dt}{1+t^{2}}= 0
\Rightarrow \tan^{-1}y+\tan^{-1}t= \tan^{-1}c
\Rightarrow \tan^{-1}y+\tan^{-1}e^{x}= \tan^{-1}c
given that y\left ( 0 \right )= 1,\tan^{-1}1+\tan^{-1}e^{0}= \tan^{-1}c
c= \tan \left ( \frac{\pi }{4}+\frac{\pi }{4} \right )= \tan \frac{\pi }{2}= \frac{1}{0}
\therefore The solution is \tan^{-1}y+\tan^{-1}e= \tan^{-1}\frac{1}{0}
ie \tan^{-1}\frac{y+e^{x}}{1-ye^{x}}= \tan^{-1}\frac{1}{0}
\Rightarrow \frac{y+e^{x}}{1-ye^{x}}= \frac{1}{0}            \therefore 1-ye^{x}= 0
\therefore y=e^{-x}  is the required solutions.
 

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