Question

\[\frac{dy}{dx} + y = \sin x\]

Answer 1

We have,
\[\frac{dy}{dx} + y = \sin x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 1\]
\[Q = \sin x \]
\[ \therefore \text{I.F.} = e^{\int P\ dx} \]
\[ = e^{\int dx} \]
\[ = e^x \]
\[\text{ Multiplying both sides of }\left( 1 \right) \text{ by }e^x , \text{ we get }\]
\[ e^x \left( \frac{dy}{dx} + y \right) = e^x \sin x\]
\[ \Rightarrow e^x \frac{dy}{dx} + e^x y = e^x \sin x \]
Integrating both sides with respect to x, we get
\[y e^x = \int e^x \sin x dx + C\]
\[ \Rightarrow y e^x = \frac{e^x}{2}\left( \sin x - \cos x \right) + C\]
\[ \Rightarrow y = C e^{- x} + \frac{1}{2}\left( \sin x - \cos x \right)\]
\[\text{Hence, } y = C e^{- x} + \frac{1}{2}\left( \sin x - \cos x \right)\text{ is the required solution.} \]