Solve y=ae^2x+be^-3x+ce^x

$$y = a e ^ { 2 x } + b e ^ { - 3 x } + c e ^ { x }$$

Solve for a

$$a=\frac{ye^{3x}-ce^{4x}-b}{e^{5x}}$$

Steps for Solving Linear Equation

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

$$ae^{2x}+be^{-3x}+ce^{x}=y$$

Subtract be^{-3x} from both sides.

$$ae^{2x}+ce^{x}=y-be^{-3x}$$

Subtract ce^{x} from both sides.

$$ae^{2x}=y-be^{-3x}-ce^{x}$$

The equation is in standard form.

$$e^{2x}a=-ce^{x}-\frac{b}{e^{3x}}+y$$

Divide both sides by e^{2x}.

$$\frac{e^{2x}a}{e^{2x}}=\frac{-ce^{x}-\frac{b}{e^{3x}}+y}{e^{2x}}$$

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

$$a=\frac{-ce^{x}-\frac{b}{e^{3x}}+y}{e^{2x}}$$

Divide -e^{x}c+y-\frac{b}{e^{3x}} by e^{2x}.

$$a=\frac{y}{e^{2x}}-\frac{b}{e^{5x}}-\frac{c}{e^{x}}$$

Solve for b

$$b=-e^{3x}\left(ce^{x}+ae^{2x}-y\right)$$

Steps for Solving Linear Equation

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

$$ae^{2x}+be^{-3x}+ce^{x}=y$$

Subtract ae^{2x} from both sides.

$$be^{-3x}+ce^{x}=y-ae^{2x}$$

Subtract ce^{x} from both sides.

$$be^{-3x}=y-ae^{2x}-ce^{x}$$

The equation is in standard form.

$$\frac{1}{e^{3x}}b=y-ae^{2x}-ce^{x}$$

Divide both sides by e^{-3x}.

$$\frac{\frac{1}{e^{3x}}be^{3x}}{1}=\frac{\left(y-ae^{2x}-ce^{x}\right)e^{3x}}{1}$$

Dividing by e^{-3x} undoes the multiplication by e^{-3x}.

$$b=\frac{\left(y-ae^{2x}-ce^{x}\right)e^{3x}}{1}$$

Divide y-ae^{2x}-ce^{x} by e^{-3x}.

$$b=e^{3x}\left(y-ae^{2x}-ce^{x}\right)$$