Question

Solve the differential equation $\frac{dy}{dx}=\frac{2x\left(\log x+1\right)}{\sin y+y\cos y}$, given that y = 0, when x = 1.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{2x\left(\log x+1\right)}{\sin y+y\cos y} \\ \Rightarrow \left(\sin y+y\cos y\right)dy=2x\left(\log x+1\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \left(\sin y+y\cos y\right)dy=\int 2x\left(\log x+1\right)dx \\ \Rightarrow \int \sin ydy+\int y\cos ydy=\int 2x\log xdx+\int 2xdx \\ \Rightarrow -\cos y+\left[y\int \cos ydy-\int \left\{\frac{d}{dy}\left(y\right)\int \cos ydy\right\}dy\right]=2\left[\log x\int xdx-\int \left\{\frac{d}{dx}\left(\log x\right)\int xdx\right\}dx\right]+x^2 \\ \Rightarrow -\cos y+\left[y\sin y+\cos y\right]=2\left[\log x\times {\frac{x}{2}}^2-\frac{x^2}{4}\right]+x^2+C \\ \Rightarrow y\sin y=x^2\log x-\frac{x^2}{2}+x^2+C \\ \Rightarrow y\sin y=x^2\log x+\frac{x^2}{2}+C\mathit{}.....\left(1\right) \\ \mathrm{Given}:x=1,y=0. \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 0=0+\frac{1}{2}+C \\ \Rightarrow C=-\frac{1}{2} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y\sin y=x^2\log x+\frac{x^2}{2}-\frac{1}{2} \\ \Rightarrow 2y\sin y=2x^2\log x+x^2-1 \\ \mathrm{Hence},2y\sin y=2x^2\log x+x^2-1\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$