Solve e^2x+y+ce^y=1

$$e ^ { 2 x + y } + c e ^ { y } = 1$$

Solve for c

$$c=\frac{1}{e^{y}}-e^{2x}$$

Steps for Solving Linear Equation

Subtract e^{2x+y} from both sides.

$$ce^{y}=1-e^{2x+y}$$

The equation is in standard form.

$$e^{y}c=1-e^{y+2x}$$

Divide both sides by e^{y}.

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

Dividing by e^{y} undoes the multiplication by e^{y}.

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

Solve for x (complex solution)

$$x=\frac{\ln(1-ce^{y})-y}{2}+\pi n_{1}i$$

$$n_{1}\in \mathrm{Z}$$

$$c\neq \frac{1}{e^{y}}$$

Solve for x

$$x=\frac{\ln(1-ce^{y})-y}{2}$$

$$c<\frac{1}{e^{y}}$$