Question

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

Answer 1

We have,

\[\frac{dy}{dx} + y = 4x . . . . . \left( 1 \right)\]

Clearly, it is a linear differential equation of the form

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

\[\text{where }P = 1\text{ and }Q = 4x\]

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

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

\[ = e^x \]

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

\[ e^x \left( \frac{dy}{dx} + y \right) = e^x 4x \]

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

Integrating both sides with respect to x, we get

\[y e^x = 4\int x e^x dx + C\]

\[ \Rightarrow y e^x = 4x\int e^x dx - 4\int\left[ \frac{d}{dx}\left( x \right)\int e^x dx \right]dx + C\]

\[ \Rightarrow y e^x = 4x e^x - 4\int e^x dx + C\]

\[ \Rightarrow y e^x = 4x e^x - 4 e^x + C\]

\[ \Rightarrow y e^x = 4\left( x - 1 \right) e^x + C\]

\[ \Rightarrow y = 4\left( x - 1 \right) + C e^{- x} \]

\[\text{Hence, }y = 4\left( x - 1 \right) + C e^{- x}\text{ is the required solution.}\]