#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 44 Maths Textbook Solution.

Answer: $x\: y=\tan ^{-1}x+C$

Hint: you must know the rules of solving differential equation and integrations.

Given:$\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$

Solution:$\left(1+y+x^{2} y\right) d x+\left(x+x^{3}\right) d y=0$

\begin{aligned} &x\left(1+x^{2}\right) d y=-\left[\left(1+y\left(1+x^{2}\right)\right] d x\right. \\ &\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{-1-y\left(1+x^{2}\right)}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}=\frac{-1}{x} \cdot y-\frac{1}{x\left(1+x^{2}\right)} \\ &\frac{d y}{d x}+\frac{1}{x} \cdot y=-\frac{1}{x\left(1+x^{2}\right)} \end{aligned}

Comparing with  $\frac{dy}{dx}+p\: y=q$ ,  we get,

\begin{aligned} &P=\frac{1}{x} \text { and } q=-\frac{1}{x\left(1+x^{2}\right)} \\ &\text { I.F }=e^{\int \frac{1}{x} d x} \\ &=e^{\log x}=x \quad\left[\because \mathrm{e}^{\log x}=x\right] \end{aligned}

The solution is,

\begin{aligned} &\mathrm{y} \cdot \mathrm{I} \cdot \mathrm{F}=\int-\frac{1}{\mathrm{x}\left(1+\mathrm{x}^{2}\right)} \cdot \mathrm{x} \mathrm{dx} \\ &\mathrm{y} \mathrm{x}=-\int \frac{1}{\left(1+\mathrm{x}^{2}\right)} \mathrm{dx} \\ &\mathrm{yx}=-\tan ^{-1} \mathrm{x}+\mathrm{C} \end{aligned}