Question

Find the particular solution of the differential equation $\frac{dx}{dy}+x\cot y=2y+y^2$ cot y, y ≠ 0 given that x = 0 when $y=\frac{\mathrm{\pi }}{2}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dx}{dy}+x\cot y=2y+y^2\cot y.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where}P=\cot y\mathrm{and}Q=2y+y^2\cot y \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{\int \cot ydy} \\ =e^{\log \left|\sin y\right|}=\sin y \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}\mathrm{I}.\mathrm{F}.=\sin y,\mathrm{we}\mathrm{get} \\ \sin y\left(\frac{dx}{dy}+x\cot y\right)=\sin y\left(y^2\cot y+2y\right) \\ \Rightarrow \sin y\frac{dx}{dy}+x\cos y=y^2\cos y+2y\sin y \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ x\sin y=\int \underset{\mathrm{I}}{y^2}\underset{\mathrm{II}}{\cos y}dy+\int 2y\sin ydy+C \\ \Rightarrow x\sin y=y^2\int \cos ydy-\int \left[\frac{d}{dy}\left(y^2\right)\int \cos ydy\right]dy+\int 2y\sin ydy+C \\ \Rightarrow x\sin y=y^2\sin y-\int 2y\sin ydy+\int 2y\sin ydy+C \\ \Rightarrow x\sin y=y^2\sin y+C \\ \mathrm{Now}, \\ y=\frac{\mathrm{\pi }}{2}\mathrm{at}x=0 \\ \therefore 0\times \sin \frac{\mathrm{\pi }}{2}={\frac{\mathrm{\pi }}{4}}^2\sin \frac{\mathrm{\pi }}{2}+C \\ \Rightarrow C=-{\frac{\mathrm{\pi }}{4}}^2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get} \\ x\sin y=y^2\sin y-{\frac{\mathrm{\pi }}{4}}^2 \\ \mathrm{Hence},x\sin y=y^2\sin y-{\frac{\mathrm{\pi }}{4}}^2\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$