A) \[x\]
B) \[\frac{1}{x}\]
C) \[\log x\]
D) \[{{e}^{x}}\]
Correct Answer: C
Solution :
\[x\sin \left( \frac{y}{x} \right)dy=\left[ y\,\sin \left( \frac{y}{x} \right)-x \right]dx\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y\,\sin \left( \frac{y}{x} \right)-x}{x\sin \left( \frac{y}{x} \right)}=\frac{\frac{y}{x}\sin \left( \frac{y}{x} \right)-1}{\sin \left( \frac{y}{x} \right)}\] Let \[\frac{y}{x}=u\]and \[\frac{dy}{dx}=x\frac{du}{dx}+u\] \[\therefore \] \[x\frac{du}{dx}+u=\frac{u\sin u-1}{\sin u}\] \[\Rightarrow \] \[x\frac{du}{dx}=\frac{u\sin u-1-u\sin u}{\sin u}\] \[\Rightarrow \] \[-\sin udu=\frac{1}{x}dx\] On integrating both sides, we get \[\cos u=\log x+c\] \[\Rightarrow \] \[\cos \left( \frac{y}{x} \right)=\log x+c\] \[\because \] \[y\,(1)=\frac{\pi }{2}\] \[\therefore \] \[\cos \frac{\pi }{2}=\log 1+c\Rightarrow c=0\] Thus, \[\cos \left( \frac{y}{x} \right)=\log x\]
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