Solve y=c_{1}e^x+c_{2}e^-3x

$$y = c _ { 1 } e ^ { x } + c _ { 2 } e ^ { - 3 x }$$

Solve for c_1

$$c_{1}=\frac{y}{e^{x}}-\frac{c_{2}}{e^{4x}}$$

Steps for Solving Linear Equation

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

$$c_{1}e^{x}+c_{2}e^{-3x}=y$$

Subtract c_{2}e^{-3x} from both sides.

$$c_{1}e^{x}=y-c_{2}e^{-3x}$$

The equation is in standard form.

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

Divide both sides by e^{x}.

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

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

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

Divide y-\frac{c_{2}}{e^{3x}} by e^{x}.

$$c_{1}=\frac{y}{e^{x}}-\frac{c_{2}}{e^{4x}}$$

Solve for c_2

$$c_{2}=-e^{3x}\left(c_{1}e^{x}-y\right)$$

Steps for Solving Linear Equation

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

$$c_{1}e^{x}+c_{2}e^{-3x}=y$$

Subtract c_{1}e^{x} from both sides.

$$c_{2}e^{-3x}=y-c_{1}e^{x}$$

The equation is in standard form.

$$\frac{1}{e^{3x}}c_{2}=y-c_{1}e^{x}$$

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

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

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

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

Divide y-e^{x}c_{1} by e^{-3x}.

$$c_{2}=e^{3x}\left(y-c_{1}e^{x}\right)$$