A) \[x(y+\cos x)=\sin x+c\]
B) \[x(y-\cos x)=\sin x+c\]
C) \[x(y\cos x)=\sin x+c\]
D) \[x(y-\cos x)=\cos x+c\]
E) \[x(y+\cos x)=\cos x+c\]
Correct Answer: A
Solution :
\[\because \] \[\frac{dy}{dx}+\frac{y}{x}=\sin x\] Here,\[P=\frac{1}{x}\]and \[Q=\sin x\] \[\therefore \] \[IF={{e}^{\int{p}\,dx}}={{e}^{\int{1/x}\,dx}}=x\] \[\therefore \] Solution is \[y(IF)=\int{Q(IF)}\,dx+c\] \[\Rightarrow \] \[y.x=\int{x\,\sin x\,dx+c}\] \[\Rightarrow \] \[xy=-x\cos x+\sin x+c\] \[\Rightarrow \] \[x(y+\cos x)=\sin x+c\]
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