Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{xe}^{4\text{x}}$
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{xe}^{4\text{x}}$
Multiplying both sides of (1) by e2x, we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\times\text{xe}^{4\text{x}}$$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{xe}^{6\text{x}}$
Integrating both sides with respect to x, we get
$\text{e}^{2\text{x}}\text{y}=\int\text{e}^{6\text{x}}\text{xdx + C}$ $\Rightarrow\ \text{e}^{2\text{x}}\text{y}=\text{x}\int\text{e}^{6\text{x}}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{6\text{x}}\text{dx}\Big]\text{dx + C}$$\Rightarrow\ \text{e}^{2\text{x}}\text{y}=\frac{\text{x}\text{e}^{6\text{x}}}{6}-\frac{\text{e}^{6\text{x}}}{36}+\text{C}$
$\Rightarrow\ \text{y}=\frac{\text{xe}^{4\text{x}}}{6}-\frac{\text{e}^{4\text{x}}}{36}+\text{Ce}^{-2\text{x}}$ Hence, $\text{y}=\frac{\text{xe}^{4\text{x}}}{6}-\frac{\text{e}^{4\text{x}}}{36}+\text{Ce}^{-2\text{x}}$ is the required solution.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.