Question

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

Answer 1

We have,
\[\frac{dy}{dx} + y = e^{- 2x} . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = 1\]
\[Q = e^{- 2x} \]
\[ \therefore \text{I.F.} = e^{\int P\ dx} \]
\[ = e^{\int1 dx} \]
\[ = e^x \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }e^x ,\text{ we get }\]
\[ e^x \left( \frac{dy}{dx} + y \right) = e^x e^{- 2x} \]
\[ \Rightarrow e^x \frac{dy}{dx} + e^x y = e^{- x} \]
Integrating both sides with respect to x, we get
\[y e^x = \int e^{- x} dx + C\]
\[ \Rightarrow y e^x = - e^{- x} + C\]
\[ \Rightarrow y = - e^{- 2x} + C e^{- x} \]
\[\text{ Hence, }y = - e^{- 2x} + C e^{- x}\text{ is the required solution.}\]