Question

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

Answer 1

We have, 
\[\frac{dy}{dx} + 2x = e^{3x} \]
\[ \Rightarrow \frac{dy}{dx} = e^{3x} - 2x\]
\[ \Rightarrow dy = \left( e^{3x} - 2x \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( e^{3x} - 2x \right)dx\]
\[ \Rightarrow y = \frac{e^{3x}}{3} - 2\frac{x^2}{2} + C\]
\[ \Rightarrow y = \frac{e^{3x}}{3} - x^2 + C\]
\[ \Rightarrow y + x^2 = \frac{e^{3x}}{3} + C\]
\[So, y + x^2 = \frac{e^{3x}}{3} + \text{ C is defined for all }x \in R . \]
\[\text{ Hence,} y + x^2 = \frac{e^{3x}}{3} +\text{ C, where } x \in R,\text{ is the solution to the given differential equation }.\]