A) \[y=x\text{ }tan\text{ }x+sin\text{ }x+c\]
B) \[y=y\text{ }tan\text{ }x+c\]
C) \[yx\text{ }sec\text{ }x=tan\text{ }x+c\]
D) \[xy\text{ }cos\text{ }x=x+c\]
Correct Answer: D
Solution :
Given, differential equation can be rewitten as \[\frac{dy}{dx}+\left( \frac{x\sin x+\cos x}{x\cos } \right)y=\frac{1}{x\cos x}\] Here, \[P=\frac{x\sin x+\cos x}{x\cos x}\] and \[Q=\frac{1}{x\cos x}\] \[\therefore \] \[IF={{e}^{\int{P\,dx}}}\] \[={{e}^{\int{\frac{x\sin x+\cos x}{x\cos x}dx}}}\] \[={{e}^{\log (x\cos x)}}=x\cos x\] \[\therefore \]Required solution is \[y.x\cos x=\int{\frac{x\cos x}{x\cos x}}dx+c\] \[\Rightarrow \] \[xy\text{ }cos\text{ }x=x+c\]
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