#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 18 Maths Textbook Solution.

Answer: $y=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+\frac{2(x+1)^{5 / 2}}{5}-\frac{2(x+1)^{8 / 2}}{3}+C$

Hint: Apply integration by parts method.

Given:$\frac{dy}{dx}=\sin ^{3}x\cos ^{4}x+x\sqrt{x+1}$

Solution:$\frac{dy}{dx}=\sin ^{3}x\cos ^{4}x+x\sqrt{x+1}$

\begin{aligned} &\left.\Rightarrow \int d y=\int \sin ^{3} x \cos ^{4} x+x \sqrt{x+1}\right) d x \text { (integrating both sides) } \\ &\left.\Rightarrow \quad y=\int \sin ^{3} x \cos ^{4} x d x+\int x \sqrt{x+1}\right) d x \\ &\Rightarrow \quad y=I_{1}+I_{2} \end{aligned}

Now, $I_{1}=\int \sin ^{3}x\cos ^{4}xdx$

$=\int \left ( 1-\cos ^{2}x \right )\left ( \cos ^{4}x \right )\left ( \sin x \right )dx\; \; \; \; \; \; \; \; \; \; \; \left ( \because \cos ^{2}x+\sin ^{2}x=1 \right )$

Let $t=\cos x$

\begin{aligned} &\Rightarrow \quad d t=-\sin x d x(\text { diff. w.r.tx }) \\ &\Rightarrow I_{1}=-\int t^{4}\left(1-t^{2}\right) d t \\ &\left.\Rightarrow I_{1}=\int t^{4}\left(t^{2}-1\right) d t=\int t^{6}-t^{4}\right) d t \\ &\Rightarrow \quad l_{1}=\frac{t^{7}}{7}-\frac{t^{5}}{5}+c_{1} \\ &=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+c_{1} \\ &\left.l_{2}=\int x \sqrt{x+1}\right) d x \end{aligned}

Let $t^{2}=x+1$

\begin{aligned} &\Rightarrow 2 t d t=d x \quad(\text { diff. } w . r \cdot t x) \\ &\Rightarrow \quad l_{2}=2 \int\left(t^{2}-1\right) \cdot t \cdot t d t \\ &\Rightarrow \quad l_{2}=2 \int t^{4}-t^{2} d t \\ &\Rightarrow \quad I_{2}=\frac{2 t^{5}}{5}-\frac{2 t^{3}}{3}+c_{2} \end{aligned}

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

\begin{aligned} &\Rightarrow \mathrm{y}=\frac{\cos ^{7} \mathrm{x}}{7}-\frac{\cos ^{5} \mathrm{x}}{5}+\mathrm{c}_{1}+\frac{2(\mathrm{x}+1)^{5 / 2}}{5}-\frac{2(\mathrm{x}+1)^{8 / 2}}{3}+\mathrm{C}_{2} \\ &\Rightarrow \mathrm{y}=\frac{\cos ^{7} \mathrm{x}}{7}-\frac{\cos ^{5} \mathrm{x}}{5}+\frac{2(\mathrm{x}+1)^{5 / 2}}{5}-\frac{2(\mathrm{x}+1)^{8 / 2}}{3}+\mathrm{c} \quad\left(\because \mathrm{c}_{1}+\mathrm{C}_{2}=\mathrm{C}\right) \end{aligned}