#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 22 Maths Textbook Solution.

Answer: $y=\left ( x^{2}-2x +2\right )e^{x}+c$

Hint: You must know about the formula of $\int uv\; \; dx$

Given:$\frac{dy}{dx}=x^{2}e^{x}$

Solution:$\frac{dy}{dx}=x^{2}e^{x}$

\begin{aligned} &\Rightarrow d y=x^{2} e^{x} d x\\ &\Rightarrow \int d y=\int x^{2} e^{x} d x \text { (integrate both sides) }\\ &\Rightarrow \int d y=x^{2} \int e^{x} d x-\int\left(2 x \int e^{x} d x\right) d x\\ &\left[\because \int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right]\\ &\Rightarrow y=x^{2} e^{x}-2 \int x e^{x} d x\\ &\Rightarrow y=x^{2} e^{x}-2 x \int e^{x} d x+2 \int e^{x} d x\\ &\left[\because \int \mathrm{uv} \mathrm{d} \mathrm{x}=\mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{u} \int \mathrm{v} \mathrm{d} \mathrm{x}\right) \mathrm{dx}\right]\\ &\Rightarrow y=x^{2} e^{x}-2 x e^{x}+2 e^{x}+C\\ &\Rightarrow y=\left(x^{2}-2 x+2\right) e^{x}+C \end{aligned}