Question

\[\frac{dy}{dx} + 2y = \sin x\]

Answer 1

We have, 
\[\frac{dy}{dx} + 2y = \sin x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 2\]
and
\[Q = \sin x \]
\[ \therefore \text{I.F.} = e^{\int P\ dx} \]
\[ = e^{\int2 dx} \]
\[ = e^{2x} \]
\[\text{Multiplying both sides of }\left( 1 \right)\text{ by I.F.} = e^{2x} ,\text{ we get }\]
\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = e^{2x} \sin x \]
\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = e^{2x} \sin x\]
Integrating both sides with respect to x, we get
\[y e^{2x} = \int e^{2x} \sin x dx + C\]
\[ \Rightarrow y e^{2x} = \frac{1}{5}\int\left[ 2 e^{2x} \left( 2\sin x - \cos x \right) + e^{2x} \left( 2 \cos x + \sin x \right) \right] dx + C\]
\[\text{ Putting }e^{2x} \left( 2 \sin x - \cos x \right) = t\]
\[ \Rightarrow \left[ 2 e^{2x} \left( 2\sin x - \cos x \right) + e^{2x} \left( 2 \cos x + \sin x \right) \right] dx = dt\]
\[ \therefore y e^{2x} = \frac{1}{5}\int dt + C\]
\[ \Rightarrow y e^{2x} = \frac{t}{5} + C\]
\[y e^{2x} = \frac{e^{2x}}{5}\left( 2\sin x - \cos x \right) + C\]
\[ \Rightarrow y = \frac{1}{5}\left( 2\sin x - \cos x \right) + C e^{- 2x} \]
\[\text{ Hence, } y = \frac{1}{5}\left( 2\sin x - \cos x \right) + C e^{- 2x}\text{ is the required solution . }\]