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The solution of the differential equation $\frac{dy}{dx}=1+x+y^2+xy^2$ is : |
$tan^{-1}y=x+\frac{x^2}{2}+C$ $log(1+y^2)=x+\frac{x^2}{2}+C$ $tan^{-1}y=tan^{-1}x+\frac{x^2}{2}+C$ $tan^{-1}y=x+\frac{x^2}{2}+\frac{x3}{3}+C$ |
$tan^{-1}y=x+\frac{x^2}{2}+C$ |
The correct answer is Option (1) → $\tan^{-1}y=x+\frac{x^2}{2}+C$ $\frac{dy}{dx}=(1+x)+y^2+(1+x)=(1+y^2)(1+x)$ so $\int\frac{1}{1+y^2}dy=\int 1+x\,dx$ $\tan^{-1}y=\frac{x^2}{2}+x+C$ |
