A) \[x(y+\cos x)=\sin x+c\]
B) \[x(y-\cos x)=\sin x+c\]
C) \[x(y\cdot \cos x)=\sin x+c\]
D) \[x(y-\cos x)=\cos x+c\]
E) \[x(y+\cos x)=\cos x+c\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}+\frac{y}{x}=\sin x\]; I.F.\[={{e}^{\int{\frac{1}{x}dx}}}={{e}^{\log x}}=x\] \ \[yx=\int{x\sin xdx}\] Þ \[yx=\int{x\sin xdx}\] Þ \[xy=-x\cos x+\sin x+c\] Þ \[x(y+\cos x)=\sin x+c\].You need to login to perform this action.
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