Question

Solve the following differential equation:-

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

Answer 1

We have,

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

\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]

\[P = 3\]

\[Q = e^{- 2x} \]

Now,

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

\[ = e^{3\int dx} \]

\[ = e^{3x} \]

So, the solution is given by

\[y \times I . F . = \int Q \times I . F . dx + C\]

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

\[ \Rightarrow y e^{3x} = e^x + C\]

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