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
Correct Answer: A
Solution :
\[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x)}{y(2\log y+1)}\] Þ\[\int_{{}}^{{}}{(2y\log y+y)dy=\int_{{}}^{{}}{{{e}^{x}}({{\sin }^{2}}x+\sin 2x})dx}\] On integrating by parts, we get \[{{y}^{2}}(\log y)={{e}^{x}}{{\sin }^{2}}x+c\].You need to login to perform this action.
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