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 \[x=\frac{\pi }{2}.\]

Answer:

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 \[y(\sin x)={{x}^{2}}\sin x-\frac{{{\pi }^{2}}}{4}\] \[\Rightarrow \]   \[y={{x}^{2}}-\frac{{{\pi }^{2}}}{4}\cos ec\,x\]