A) \[y\sin y={{x}^{2}}\log x+c\]
B) \[y\sin y={{x}^{2}}+c\]
C) \[y\sin y={{x}^{2}}+\log x+c\]
D) \[y\sin y=x\log x+c\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}=\frac{x\log {{x}^{2}}+x}{\sin y+y\cos y}\]. Separating the variables and integrating \[\int{(\sin y+y\cos y)dy=\int{(x\log {{x}^{2}}+x)dx}}\] Þ \[-\cos y+y\sin y+\cos y\] \[=\frac{{{x}^{2}}}{2}\log {{x}^{2}}-\int{\frac{{{x}^{2}}}{2}.\frac{1}{{{x}^{2}}}.2xdx+\int{x\,dx+c}}\] Þ \[y\sin y=\frac{{{x}^{2}}}{2}2\log x-\int{x\,dx+\int{xdx+c}}\] Þ \[y\sin y={{x}^{2}}\log x+c\].
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