The solution of \[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x)}{y(2\,\,\log \,\,y+1)}\] is

\[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\cos }^{2}}x+c=0\]

\[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\sin }^{2}}x+c=0\]

\[{{y}^{2}}(\log \,y)+{{e}^{x}}{{\cos }^{2}}x+c=0\]

The solution of \[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x

1.	The solution of \[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x)}{y(2\,\,\log \,\,y+1)}\] is
A.	\[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\sin }^{2}}x+c=0\]
B.	\[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\cos }^{2}}x+c=0\]
C.	\[{{y}^{2}}(\log \,y)+{{e}^{x}}{{\cos }^{2}}x+c=0\]
D.	None of these
Answer» B. \[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\cos }^{2}}x+c=0\]

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The solution of \[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x)}{y(2\,\,\log \,\,y+1)}\] is

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