Solve the differential equation  \frac{dy}{dx}= 1+x^{2}+y^{2}+x^{2}y^{2},  given that y= 1  when x= 0

 

 

 

 
 
 
 
 

Answers (1)

\frac{dy}{dx}= 1+x^{2}+y^{2}+x^{2}y^{2}
      \Rightarrow \frac{dy}{dx}= \left ( 1+x^{2}\right )\left ( 1+y^{2} \right )
Integrating
              \Rightarrow \int \frac{dy}{1+y^{2}}= \int \left ( 1+x^{2} \right )dx
             \Rightarrow \tan ^{-1}y= x+\frac{x^{3}}{3}+c
As y= 1, when  x= 0  so \tan ^{-1}1= 0+0+c\Rightarrow c= \frac{\pi }{4}
Hence, the required solution is \tan ^{-1}y= x+\frac{x^{3}}{3}+\frac{\pi }{4}
               or y=\tan \left ( x+\frac{x^{3}}{3}+\frac{\pi }{4} \right ).

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