Solve y e ^ 2 x = ∫ e ^2x(xe^x+c_{1}e^x) wrt x

$$y e ^ { 2 x } = \int e ^ { 2 x } ( x e ^ { x } + c _ { 1 } e ^ { x } ) d x$$

Solve for y

$$y=\frac{c_{1}e^{x}}{3}+\frac{xe^{x}}{3}+\frac{С}{e^{2x}}-\frac{e^{x}}{9}$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{2x} by xe^{x}+c_{1}e^{x}.

$$ye^{2x}=\int e^{2x}xe^{x}+e^{2x}c_{1}e^{x}\mathrm{d}x$$

The equation is in standard form.

$$e^{2x}y=\frac{c_{1}e^{3x}}{3}+\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+С$$

Divide both sides by e^{2x}.

$$\frac{e^{2x}y}{e^{2x}}=\frac{\frac{c_{1}e^{3x}}{3}+\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+С}{e^{2x}}$$

Dividing by e^{2x} undoes the multiplication by e^{2x}.

$$y=\frac{\frac{c_{1}e^{3x}}{3}+\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}+С}{e^{2x}}$$

Divide \frac{e^{3x}x}{3}-\frac{e^{3x}}{9}+\frac{c_{1}e^{3x}}{3}+С by e^{2x}.

$$y=\frac{c_{1}e^{x}}{3}+\frac{xe^{x}}{3}+\frac{С}{e^{2x}}-\frac{e^{x}}{9}$$

Solve for c_1

$$\left\{\begin{matrix}c_{1}=\frac{С}{xe^{3x}}+\frac{y}{xe^{x}}-x\text{, }&x\neq 0\\c_{1}\in \mathrm{R}\text{, }&y=С\text{ and }x=0\end{matrix}\right.$$