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Question
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
Options
x (y + cos x) = sin x + C
x (y − cos x) = sin x + C
x (y + cos x) = cos x + C
none of these
Solution
x (y + cos x) = sin x + C
We have,
\[\frac{dy}{dx} + \frac{y}{x} = \sin x\]
\[ \Rightarrow \frac{dy}{dx} + \frac{1}{x}y = \sin x . . . . . \left( 1 \right)\]
\[\text{ Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]
\[P = \frac{1}{x} \]
\[Q = \sin x\]
Now,
\[I . F . = e^{\int\frac{1}{x}dx} = e^{log\left| x \right|} \]
\[ = x\]
\[\text{ Therefore, integration of }\left( 1 \right) \text{ is given by }\]
\[y \times I . F . = \int x^2 \times I . F . dx + C\]
\[ \Rightarrow yx = x\int\sin x dx - \int\left[ \frac{d}{dx}\left( x \right)\int\sin x dx \right]dx + C\]
\[ \Rightarrow yx = - x \cos x + \int\cos x dx + C\]
\[ \Rightarrow yx + x \cos x = \sin x + C\]
\[ \Rightarrow x\left( y + \cos x \right) = \sin x + C\]
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