Question

The solution of the differential equation $\frac{dy}{dx}=1+x+y^2+x{y}^2,y\left(0\right)=0$ is
(a) $y^2=\exp \left(x+\frac{x^2}{2}\right)-1$

(b) $y^2=1+C\exp \left(x+\frac{x^2}{2}\right)$

(c) y = tan (C + x + x²)

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

Answer 1

$\left(\mathrm{d}\right)y=\tan \left(x+\frac{x^2}{2}\right)$

We have,
$$\begin{gathered} \frac{dy}{dx}=1+x+y^2+x{y}^2 \\ \Rightarrow \frac{dy}{dx}=\left(x+1\right)+y^2\left(x+1\right) \\ \Rightarrow \frac{dy}{dx}=\left(x+1\right)\left(1+y^2\right) \\ \Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(x+1\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{dy}{\left(1+y^2\right)}=\int \left(x+1\right)dx \\ \Rightarrow {\tan }^{-1}y=\frac{x^2}{2}+x+C.....\left(1\right) \\ \mathrm{Now}, \\ y\left(0\right)=0 \\ \therefore {\tan }^{-1}0=\frac{0}{2}+0+C \\ \Rightarrow C=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {\tan }^{-1}y=\frac{x^2}{2}+x \\ \Rightarrow y=\tan \left(\frac{x^2}{2}+x\right) \end{gathered}$$