#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 45 sub question (vi)

Answer: $\tan ^{-1} y=\frac{x^{3}}{3}+x+\frac{\pi}{4}$

Hint: Separate the terms of x and y and then integrate them.

Given:  $\frac{d y}{d x}=1+x^{2}+y^{2}+x^{2} y^{2}, y(0)=1$

Solution:

\begin{aligned} &\frac{d y}{d x}=1+x^{2}+y^{2}+x^{2} y^{2} \\\\ &\Rightarrow \frac{d y}{d x}=\left(1+x^{2}\right)+y^{2}\left(1+x^{2}\right) \\\\ &\Rightarrow \frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right) \\\\ &\Rightarrow \frac{d y}{1+y^{2}}=\left(1+x^{2}\right) d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int\left(1+x^{2}\right) d x \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+c \end{aligned}        ...............(1)

Given that $y(0)=1 \text { i.e. at } x=0, y=1$

\begin{aligned} &\tan ^{-1}(1)=0+\frac{0}{3}+c \Rightarrow c=\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\\\ &{\left[\therefore \tan \frac{\pi}{4}=1\right]} \\\\ &\Rightarrow c=\frac{\pi}{4} \\\\ &\tan ^{-1} y=x+\frac{x^{3}}{3}+\frac{\pi}{4} \end{aligned}