Question

Solve the following initial value problem:-

\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]

Answer 1

We have, 
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = 2\text{ and }Q = e^{- 2x} \sin x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int2 dx} \]
\[ = e^{2x} \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = e^{2x} ,\text{ we get }\]
\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = e^{2x} e^{- 2x} \sin x\]
\[ \Rightarrow e^{2x} \left( \frac{dy}{dx} + 2y \right) = \sin x\]
Integrating both sides with respect to x, we get
\[y e^{2x} = \int\sin x dx + C\]
\[ \Rightarrow y e^{2x} = - \cos x + C . . . . . \left( 2 \right)\]
Now, 
\[y\left( 0 \right) = 0\]
\[ \therefore 0 \times e^0 = - \cos 0 + C\]
\[ \Rightarrow C = 1\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y e^{2x} = - \cos x + 1\]
\[ \Rightarrow y e^{2x} = 1 - \cos x\]
\[\text{ Hence, }y e^{2x} = 1 - \cos x\text{ is the required solution.}\]