Question

The general solution of the differential equation $\frac{dy}{dx}+y$ cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C

Answer 1

(d) y sin x = x + C


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+y\cot x=\mathrm{cosec}x \\ \frac{dy}{dx}+y\cot x=\mathrm{cosec}x \\ \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=\cot x \\ Q=\mathrm{cosec}x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \cot xdx}=e^{\log \left(\sin x\right)} \\ =\sin x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\sin x=\int \sin x\times \mathrm{cosec}xdx+C \\ \Rightarrow y\sin x=x+C \end{gathered}$$