Question

The particular solution of the differential equation $ \frac{dy}{dx}+y\,\cot \,x=2x+{{x}^{2}}\,\cot \,x, $ such that $ y(\pi /2)=0 $ is

• A. $ y=\frac{{{\pi }^{2}}}{4\,\cos \,x},(x\ne 0) $
• B. $ y={{x}^{2}}-\frac{\pi }{2}\tan x $
• C. $ y=\frac{2x}{\sin x}+\frac{1}{{{x}^{2}}},(x\ne 0) $
• D. $ y={{x}^{2}}-\frac{{{\pi }^{2}}}{4\sin x}(\sin x\ne 0) $

Answer 1

Answer: (D)

Given, $ \frac{dy}{dx}+y\cot x=2x+{{x}^{2}}\cot x $
$ IF={{e}^{\int{\cot \,x\,dx}}}={{e}^{\log \,\sin x}}=\sin \,x $
$ \therefore $ Complete solution is $ y.\sin x=\int{\sin x(2x+{{x}^{2}}\,\cot \,x)\,dx+C} $
$ \Rightarrow $ $ y.\sin \,x=2\int{x\,\sin \,x\,dx+\int{{{x}^{2}}\,\cos \,x\,dx+C}} $
$ \Rightarrow $ $ y.\sin \,x=2\int{x\,\sin \,x\,dx+{{x}^{2}}\,\sin x} $ $ -2\int{x\,\sin \,x\,dx+C} $
$ \Rightarrow $ $ y\,\sin \,x={{x}^{2}}\,\sin \,x+C $ ..(i)
$ \because $ $ y\left( \frac{\pi }{2} \right)=0 $ i.e, at $ x=\frac{\pi }{2},\,\,\,\,\,y=0 $
Then from Eq. (i). $ 0.\,\,\sin \,\frac{\pi }{2}={{\left( \frac{\pi }{2} \right)}^{2}}\sin \frac{\pi }{2}+C $
$ \Rightarrow $ $ C=-\frac{{{\pi }^{2}}}{4} $
Putting the value of C in Eq. (i), we get
$ y.\sin x={{x}^{2}}\sin x-\frac{{{\pi }^{2}}}{4}. $
$ \Rightarrow $ $ y={{x}^{2}}-\frac{{{\pi }^{2}}}{4\sin x} $
$ (\sin x\ne 0) $