Solve y = e ^ - ∫ wrt x[inte^-x*e^intdxdx+c]

$$y = e ^ { - \int d x } [ \int e ^ { - x } \cdot e ^ { \int d x } d x + c ]$$

Solve for c

$$c=-xe^{С_{1}}+ye^{x+С}-С_{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{-\int 1\mathrm{d}x} by \int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x+c.

$$y=e^{-\int 1\mathrm{d}x}\int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x+e^{-\int 1\mathrm{d}x}c$$

Swap sides so that all variable terms are on the left hand side.

$$e^{-\int 1\mathrm{d}x}\int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x+e^{-\int 1\mathrm{d}x}c=y$$

Subtract e^{-\int 1\mathrm{d}x}\int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x from both sides.

$$e^{-\int 1\mathrm{d}x}c=y-e^{-\int 1\mathrm{d}x}\int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x$$

Reorder the terms.

$$ce^{-\int 1\mathrm{d}x}=-e^{-\int 1\mathrm{d}x}\int e^{-x}e^{\int 1\mathrm{d}x}\mathrm{d}x+y$$

The equation is in standard form.

$$\frac{1}{e^{x+С}}c=-\frac{xe^{С}+С_{1}}{e^{x+С_{2}}}+y$$

Divide both sides by e^{-\left(x+С\right)}.

$$\frac{\frac{1}{e^{x+С}}ce^{x+С}}{1}=\frac{-xe^{С_{1}}+ye^{x+С}-С_{2}}{e^{x+С_{3}}\times \frac{1}{e^{x+С}}}$$

Dividing by e^{-\left(x+С\right)} undoes the multiplication by e^{-\left(x+С\right)}.

$$c=\frac{-xe^{С_{1}}+ye^{x+С}-С_{2}}{e^{x+С_{3}}\times \frac{1}{e^{x+С}}}$$

Divide \frac{ye^{x+С}-e^{С}x-С}{e^{x+С}} by e^{-\left(x+С\right)}.

$$c=-xe^{С_{1}}+ye^{x+С}-С_{2}$$

Solve for y

$$y=\frac{xe^{С}+c+С_{1}}{e^{x+С_{2}}}$$