Question

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

Answer 1

We have, 

\[\frac{dy}{dx} + 2y = e^{3x} . . . . . (1)\]

Clearly, it is a linear differential equation of the form 

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

where

\[P = 2\]

\[Q = e^{3x} \]

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

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

\[ = e^{2x} \]

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

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

\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = e^{5x} \]

Integrating both sides with respect to x, we get

\[y e^{2x} = \int e^{5x} dx + C\]

\[ \Rightarrow y e^{2x} = \frac{e^{5x}}{5} + C\]

\[ \Rightarrow y = \frac{1}{5} e^{3x} + C e^{- 2x} \]

\[\text{ Hence, }y = \frac{1}{5} e^{3x} + C e^{- 2x}\text{ is the required solution .} \]