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) \[y={{x}^{2}}-\frac{{{\pi }^{2}}}{4\sin x}(\sin x\ne 0)\]
Correct Answer: D
Solution :
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 \[y.\sin x={{x}^{2}}\sin x-\frac{{{\pi }^{2}}}{4}.\] \[\Rightarrow \] \[y={{x}^{2}}-\frac{{{\pi }^{2}}}{4\sin x}\] \[(\sin x\ne 0)\]
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