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