• # question_answer Find the particular solution of the differential equation $\frac{dy}{dx}+y\cot x=2x+{{x}^{2}}\cot x(x\ne 0),$ given that $y=0$ when $x=\frac{\pi }{2}.$

We have, $\frac{dy}{dx}+y\cot x=2x+{{x}^{2}}\cot x\,\,(x\ne 0)$ On comparing with the form$\frac{dy}{dx}+Py=Q$, we get $P=\cot x,\,\,Q=2x+{{x}^{2}}\cot x$ $\therefore$$lF={{e}^{\int{pdx}}}={{e}^{\int{\cot xdx}}}={{e}^{\log |\sin x|}}=\sin x$ $\therefore$The general solution of the given differential equation is given by $y\cdot lF=\int{Q\cdot lFdx+C}$ $\Rightarrow$   $y\,(\sin x)=\int{\sin x\,(2x+{{x}^{2}}\cot x)\,\,dx}$ $=\int{(2x\sin x+{{x}^{2}}\cos x)dx}$ $\left[ \because \cot x=\frac{\cos x}{\sin x} \right]$ $=\int{2x\sin xdx+\int{{{x}^{2}}\cos xdx}}$ $=\int{2x\sin xdx+\left[ {{x}^{2}}\int{\cos xdx-\int{\left\{ \frac{d{{x}^{2}}}{dx}\int{\cos xdx} \right\}dx}} \right]}$[using integration by parts in 2nd integral] $\Rightarrow$$y\sin x=\int{2x\sin xdx+{{x}^{2}}\sin x-\int{2x\sin xdx}}$ $={{x}^{2}}\sin x+C$                                       ?(i) $\therefore$At $x=\frac{\pi }{2},$ $y=0$ From Eq. (i), we get $0\sin \left( \frac{\pi }{2} \right)={{\left( \frac{\pi }{2} \right)}^{2}}\sin \frac{\pi }{2}+C$ $\Rightarrow$   $0=\frac{{{\pi }^{2}}}{4}\times 1+C$$\Rightarrow$$C=\frac{-{{\pi }^{2}}}{4}$ Hence, the particular solution is $y(\sin x)={{x}^{2}}\sin x-\frac{{{\pi }^{2}}}{4}$ $\Rightarrow$   $y={{x}^{2}}-\frac{{{\pi }^{2}}}{4}\cos ec\,x$