Question

$\frac{dy}{dx}={\cos }^{3}x{\sin }^{2}x+x\sqrt{2x+1}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}={\cos }^{3}x{\sin }^{2}x+x\sqrt{2x+1} \\ \Rightarrow dy=\left({\cos }^{3}x{\sin }^{2}x+x\sqrt{2x+1}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left({\cos }^{3}x{\sin }^{2}x+x\sqrt{2x+1}\right)dx \\ \Rightarrow y=\int {\cos }^{3}x{\sin }^{2}xdx+\int x\sqrt{2x+1}dx \\ \Rightarrow y=I_1+I_2.....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{where} \\ {I}_1=\int {\cos }^{3}x{\sin }^{2}xdx \\ {I}_2=\int x\sqrt{2x+1}dx \\ \mathrm{Now}, \\ {I}_1=\int {\cos }^{3}x{\sin }^{2}xdx \\ =\int {\sin }^{2}x\left(1-{\sin }^{2}x\right)\cos xdx \\ \mathrm{Putting}t=\sin x,\mathrm{we}\mathrm{get} \\ dt=\cos xdx \\ \Rightarrow I_1=\int t^2\left(1-t^2\right)dt \\ =\int \left(t^2-t^4\right)dt \\ =\frac{t^3}{3}-\frac{t^5}{5}+C_1 \\ =\frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C_1 \\ {I}_2=\int x\sqrt{2x+1}dx \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}{t}^2=2x+1,\mathrm{we}\mathrm{get} \\ 2tdt=2dx \\ \Rightarrow tdt=dx \\ \mathrm{Now}, \\ {I}_2=\int \left(\frac{t^2-1}{2}\right)t\times tdt \\ =\frac{1}{2}\int \left(t^4-t^2\right)dt \\ =\frac{t^5}{10}-\frac{t^3}{6}+C_2 \\ =\frac{{\left(2x+1\right)}^{{\displaystyle \frac{5}{2}}}}{10}-\frac{{\left(2x+1\right)}^{\frac{3}{2}}}{6}+C_2 \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}{I}_1\mathrm{and}{I}_2\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=\frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C_1+\frac{{\left(2x+1\right)}^{{\displaystyle \frac{5}{2}}}}{10}-\frac{{\left(2x+1\right)}^{\frac{3}{2}}}{6}+C_2 \\ y=\frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+\frac{{\left(2x+1\right)}^{{\displaystyle \frac{5}{2}}}}{10}-\frac{{\left(2x+1\right)}^{\frac{3}{2}}}{6}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathrm{Where},C=C_1+C_2\right) \\ \mathrm{Hence},y=\frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+\frac{{\left(2x+1\right)}^{{\displaystyle \frac{5}{2}}}}{10}-\frac{{\left(2x+1\right)}^{\frac{3}{2}}}{6}+C\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$