Answer:
Given differential equation is \[\frac{dy}{dx}=\frac{x(2\log \,x+1)}{\sin y+y\cos y'}\] On separating the variables, we get \[(\sin y+y\cos y)dy=(2\log \,x+x)dx\] On integrating both sides, we get \[\int{(\sin y+y\cos y)dy}=\int{(2\log \,x+x)dx}\] \[\int{\sin y\,dy+\int{\underset{I}{\mathop{y}}\,\cos \underset{II}{\mathop{y}}\,\,dy}}=2\int{\underset{II}{\mathop{x}}\,\underset{I}{\mathop{\log }}\,\,x\,dx+\int{x\,dx}}\] \[\Rightarrow \] \[-\,\cos y+[y\cdot \sin \,y-\int{1\cdot \sin y\,dy}]\] \[=2\left[ \log \,x\cdot \frac{{{x}^{2}}}{2}-\int{\frac{1}{x}\cdot \frac{{{x}^{2}}}{2}dx} \right]+\frac{{{x}^{2}}}{2}\] \[\Rightarrow \] \[-\cos \,y+y\sin \,y+\cos \,y={{x}^{2}}\cdot \log \,x-\int{x\,dx}+\frac{{{x}^{2}}}{2}\] \[\Rightarrow \] \[y\sin \,y={{x}^{2}}\log x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{2}}}{2}+C\] \[\Rightarrow \] \[y\sin \,y={{x}^{2}}\log x+C\] ? (i) Since, it is given that \[y=\frac{\pi }{2},\] when x = 1. \[\therefore \] From Eq. (i), we get \[\frac{\pi }{2}\cdot \sin \frac{\pi }{2}={{(1)}^{2}}\cdot \log \,(1)+C\] \[\Rightarrow \] \[\frac{\pi }{2}\times (1)=1\times 0+C\] \[\left[ \because \,\,\sin \frac{\pi }{2}=1\,\,\text{and}\,\,\log 1=0 \right]\] \[\Rightarrow \] \[C=\frac{\pi }{2}\] Hence, the required particular solution is \[y\sin \,y={{x}^{2}}\log \,x=\frac{\pi }{2}.\]
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