Question

Solution of the differential equation $\frac{dy}{dx}+\frac{y}{x}$ = sin x is
(a) x (y + cos x) = sin x + C
(b) x (y − cos x) = sin x + C
(c) x (y + cos x) = cos x + C
(d) none of these

Answer 1

(a) x (y + cos x) = sin x + C


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\frac{y}{x}=\sin x \\ \Rightarrow \frac{dy}{dx}+\frac{1}{x}y=\sin x.....\left(1\right) \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{x} \\ Q=\sin x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{x}dx}=e^{\log \left|x\right|} \\ =x \\ \mathrm{Therefore},\mathrm{integration}\mathrm{of}\left(1\right)\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int x^{\mathit{2}}\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow yx=\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{\sin x}dx+C \\ \Rightarrow yx=x\int \sin xdx-\int \left[\frac{d}{dx}\left(x\right)\int \sin xdx\right]dx+C \\ \Rightarrow yx=-x\cos x+\int \cos xdx+C \\ \Rightarrow yx+x\cos x=\sin x+C \\ \Rightarrow x\left(y+\cos x\right)=\sin x+C \end{gathered}$$