Question

\[4\frac{dy}{dx} + 8y = 5 e^{- 3x}\]

Answer 1

We have,
\[4\frac{dy}{dx} + 8y = 5 e^{- 3x}\]
\[\Rightarrow \frac{dy}{dx} + 2y = \frac{5}{4} e^{- 3x} . . . . . (1)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 2\]
\[Q = \frac{5}{4} e^{- 3x} \]
\[ \therefore \text{I. F.}= e^{\int P dx} \]
\[ = e^{\int2 dx} \]
\[ = e^{2x} \]
\[\text{ Multiplying both sides of (1) by }e^{2x} ,\text{ we get }\]
\[ e^{2x} \left( \frac{dy}{dx} + 2y \right) = \frac{5}{4} e^{2x} e^{- 3x} \]
\[ \Rightarrow e^{2x} \frac{dy}{dx} + 2 e^{2x} y = \frac{5}{4} e^{- x} \]
Integrating both sides with respect to x, we get
\[y e^{2x} = \frac{5}{4}\int e^{- x} dx + C\]
\[ \Rightarrow y e^{2x} = - \frac{5}{4} e^{- x} + C\]
\[ \Rightarrow y = \frac{5}{4} e^{- 3x} + C e^{- 2x} \]
\[\text{ Hence, }y = \frac{5}{4} e^{- 3x} + C e^{- 2x}\text{ is the required solution .} \]