A) \[xy\sin x=1-\cos x\]
B) \[xy\sin x+\cos x=0\]
C) \[x\sin x+y\cos x=0\]
D) \[x\sin x+y\cos x=1\]
Correct Answer: A
Solution :
[a] The equation is \[\frac{dy}{dx}+\left( \frac{x\cos x+\sin x}{x\sin x} \right)y=\frac{1}{x}\] Integrating factor I.F. \[={{e}^{\int{\frac{x\cos x+\sin x}{x\sin x}dx}}}={{e}^{\log (xsinx)}}=x\sin x\] \[\therefore \] The solution is \[y(xsinx)=\int{\frac{1}{x}(xsinx)dx+c}\] \[xy\sin x=-\cos x+c\] when \[x=0,y=0\Rightarrow c=\cos 0=1\] \[\therefore \] The particular solution is \[xy\sin x=1-\cos x\]You need to login to perform this action.
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