Question

\[\frac{dy}{dx} + 2y = x e^{4x}\]

Answer 1

We have, 
\[\frac{dy}{dx} + 2y = x e^{4x} . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 2\]
\[Q = x e^{4x} \]
\[ \therefore I.F. = e^{\int P\ dx} \]
\[ = e^{ \int2dx } \]
\[ = e^{2x} \]
\[\text{Multiplying both sides of }\left( 1 \right)\text{ by }e^{2x} ,\text{ we get }\]
\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = e^{2x} \times x e^{4x} \]
\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = x e^{6x} \]
Integrating both sides with respect to x, we get

\[ \Rightarrow e^{2x} y = x\int e^{6x} dx - \int\left[ \frac{d}{dx}\left( x \right)\int e^{6x} dx \right]dx + C\]
\[ \Rightarrow e^{2x} y = \frac{x e^{6x}}{6} - \frac{e^{6x}}{36} + C\]
\[ \Rightarrow y = \frac{x e^{4x}}{6} - \frac{e^{4x}}{36} + C e^{- 2x} \]
\[\text{ Hence, }y = \frac{x e^{4x}}{6} - \frac{e^{4x}}{36} + C e^{- 2x}\text{ is the required solution.}\]