Question

\[\frac{dy}{dx} + 5y = \cos 4x\]

Answer 1

We have,

\[\frac{dy}{dx} + 5y = \cos 4x . . . . . \left( 1 \right)\]

Clearly, it is a linear differential equation of the form

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

\[\text{where }P = 15\text{ and }Q = \cos 4x\]

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

\[ = e^{\int 5dx} \]

\[ = e^{5x} \]

\[\text{Multiplying both sides of (1) by }I.F. = e^{5x},\text{ we get}\]

\[e^{5x} \left( \frac{dy}{dx} + 5y \right) = e^{5x} \cos 4x \]

\[\Rightarrow e^{5x} \frac{dy}{dx} + 5 e^{5x} y = e^{5x} \cos 4x\]

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

\[y e^{5x} = \int e^{5x} \cos 4x dx + C\]

\[ \Rightarrow y e^{5x} = I + C . . . . . \left( 2 \right)\]

Where,

\[I = \int e^{5x} \cos 4x dx . . . . . \left( 3 \right)\]

\[ \Rightarrow I = e^{5x} \int\cos 4x dx - \int\left[ \frac{d e^{5x}}{dx}\int\cos 4x dx \right]dx\]

\[ \Rightarrow I = \frac{e^{5x} \sin 4x}{4} - \frac{5}{4}\int e^{5x} \sin 4x dx\]

\[ \Rightarrow I = \frac{e^{5x} \sin 4x}{4} - \frac{5}{4}\left[ e^{5x} \int\sin 4x dx - \int\left( \frac{d e^{5x}}{dx}\int\sin 4x dx \right)dx \right]\]

\[ \Rightarrow I = \frac{e^{5x} \sin 4x}{4} - \frac{5}{4}\left[ - \frac{e^{5x} \cos 4x}{4} + \frac{5}{4}\int e^{5x} \cos 4x dx \right]\]

\[ \Rightarrow I = \frac{e^{5x} \sin 4x}{4} + \frac{5 e^{5x} \cos 4x}{16} - \frac{25}{16}\int e^{5x} \cos 4x dx\]

\[ \Rightarrow I = \frac{e^{5x} \sin 4x}{4} + \frac{5 e^{5x} \cos 4x}{16} - \frac{25}{16}I ............\left[\text{From (3)} \right]\]

\[ \Rightarrow \frac{41}{16}I = \frac{e^{5x} \sin 4x}{4} + \frac{5 e^{5x} \cos 4x}{16}\]

\[ \Rightarrow \frac{41}{16}I = \frac{e^{5x}}{16}\left( 4\sin 4x + 5\cos 4x \right)\]

\[ \Rightarrow I = \frac{e^{5x}}{41}\left( 4\sin 4x + 5\cos 4x \right) . . . . . . . . \left( 4 \right)\]

From (2) and (4) we get

\[ \Rightarrow y e^{5x} = \frac{e^{5x}}{41}\left( 4\sin 4x + 5\cos 4x \right) + C\]

\[ \Rightarrow y = \frac{4}{41}\left( \sin 4x + \frac{5}{4}\cos 4x \right) + C e^{- 5x} \]

\[\text{Hence, }y = \frac{4}{41}\left(\sin 4x + \frac{5}{4}\cos 4x \right) + C e^{- 5x}\text{ is the required solution.}\]