Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 25 Maths Textbook Solution.

Answer:$\frac{\cos ^{5}}{5}-\frac{\cos ^{3}x}{3}+\left ( x-1 \right )e^{x}+C$

Hint:  Apply integration by parts method and formula of$\int u\; v\: dx$

Given:$\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}$

Solution:$\frac{dy}{dx}=\sin ^{3}x\cos ^{2}x+xe^{x}$

$\Rightarrow \int dy=\int \left ( \sin ^{3}x\cos ^{2}x+xe^{x} \right )dx$        (integrate both sides)

$\Rightarrow \int y=\int \sin ^{3}x\cos ^{2}xdx+\int xe^{x}dx$

$\Rightarrow y=I_{1}+I_{2}$

$I_{1}=\int \sin ^{3}x\cos ^{2}xdx$

$=\int \left ( 1-\cos ^{2}x \right )\cos ^{2}x\sin xdx$

Let $\cos x=t$

$\Rightarrow d \mathrm{t}=-\sin \mathrm{x} \mathrm{d} \mathrm{x} \quad \text { (diff. w.r.t. } \mathrm{x})$

$I_{1}=-\int t^{2}\left ( 1-t^{2} \right )dt$

$=\int \mathrm{t}^{2}\left(\mathrm{t}^{2}-1\right) \mathrm{dt}=\int \mathrm{t}^{4}-\mathrm{t}^{2} \mathrm{dt} \\$

$\Rightarrow \mathrm{I}_{1}=\frac{\mathrm{t}^{5}}{5}-\frac{\mathrm{t}^{8}}{3}+\mathrm{c}_{1} \\$

$\Rightarrow \mathrm{I}_{1}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}$

Now,$I_{2}=\int xe^{x}dx$

\begin{aligned} &\Rightarrow \mathrm{I}_{2}=x \int \mathrm{e}^{x} \mathrm{dx}-\int \mathrm{e}^{\mathrm{x}} \mathrm{dx} \\ &{\left[\because \int \mathrm{uvd} \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 \mathrm{I}_{2}=x \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)+\mathrm{c}_{2} \\ &\text { Now, } \mathrm{y}=\mathrm{I}_{1}+\mathrm{I}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+\mathrm{c}_{1}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{5} \mathrm{x}}{5}-\frac{\cos ^{8} \mathrm{x}}{3}+(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{c}\left(\because \mathrm{c}_{1}+\mathrm{c}_{2}=\mathrm{c}\right) \end{aligned}