Question

\[\frac{dy}{dx} + 2y = \sin 3x\]

Answer 1

We have,

\[\frac{dy}{dx} + 2y = \sin 3x . . . . . \left( 1 \right)\]

Clearly, it is a linear differential equation of the form

\[\frac{dy}{dx} + Py = Q\]

\[\text{where }P = 2\text{ and }Q = \sin 3x\]

\[ \therefore I . F . = e^{\int P\ dx} \]

\[ = e^{\int2 dx} \]

\[ = e^{2x} \]

\[\text{Multiplying both sides of (1) by }I.F. = e^{2x},\text{ we get}\]

\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = e^{2x} \sin 3x \]

\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = e^{2x} \sin 3x\]

Integrating both sides with respect to `x`, we get

\[y e^{2x} = \int e^{2x} \sin 3x dx + C\]

\[ \Rightarrow y e^{2x} = I + C . . . . . \left( 1 \right)\]

\[\text{Where, }I = \int e^{2x} \sin 3x dx . . . . . \left( 2 \right)\]

\[ \Rightarrow I = e^{2x} \int\sin 3x dx - \int\left[ \frac{d e^{2x}}{dx}\int\sin 3x dx \right]dx\]

\[ \Rightarrow I = - \frac{e^{2x} \cos 3x}{3} + \frac{2}{3}\int e^{2x} \cos 3x dx\]

\[ \Rightarrow I = - \frac{e^{2x} \cos 3x}{3} + \frac{2}{3}\left[ e^{2x} \int\cos 3x dx - \int\left( \frac{d e^{2x}}{dx}\int\cos 3x dx \right)dx \right]\]

\[ \Rightarrow I = - \frac{e^{2x} \cos 3x}{3} + \frac{2}{3}\left[ \frac{e^{2x} \sin 3x}{3} - \frac{2}{3}\int e^{2x} \sin 3x dx \right]\]

\[ \Rightarrow I = - \frac{e^{2x} \cos 3x}{3} + \frac{2 e^{2x} \sin 3x}{9} - \frac{4}{9}\int e^{2x} \sin 3x dx\]

\[ \Rightarrow I = - \frac{e^{2x} \cos 3x}{3} + \frac{2 e^{2x} \sin 3x}{9} - \frac{4}{9}I ............\left[\text{Using (2)} \right]\]

\[ \Rightarrow \frac{13I}{9} = - \frac{e^{2x} \cos 3x}{3} + \frac{2 e^{2x} \sin 3x}{9}\]

\[ \Rightarrow I = \frac{9}{13}\left( \frac{2 e^{2x} \sin 3x}{9} - \frac{e^{2x} \cos 3x}{3} \right)\]

\[ \Rightarrow I = \frac{e^{2x}}{13}\left( 2 \sin 3x - 3 \cos 3x \right) . . . . . \left( 3 \right)\]

From (1) and (3), we get

\[y e^{2x} = \frac{e^{2x}}{13}\left( 2 \sin 3x - 3 \cos 3x \right) + C\]

\[ \Rightarrow y = \frac{3}{13}\left( \frac{2}{3}\sin 3x - \cos 3x \right) + C e^{- 2x} \]

\[\text{Hence, }y = \frac{3}{13}\left( \frac{2}{3}\sin 3x - \cos 3x \right) + C e^{- 2x}\text{ is the required solution.}\]