Question

Show that the solution of the differential equation $\frac{dy}{dx}=1+x{y}^2+x+y^2,y\left(0\right)=0$ is $y=\tan (x+\frac{x^2}{2})$.

Answer 1

$y=\tan (x+\frac{x^2}{2})$

differentiating on both sides, we get,

$\frac{dy}{dx}={\sec }^2(x+\frac{x^2}{2})(1+x)$

$=[1+{\tan }^2(x+\frac{x^2}{2})](1+x)$

$=(1+y^2)(1+x)$

$\therefore \frac{dy}{dx}=1+x{y}^2+x+y^2$

Hence proved.