Two-parameter family of solutions of the second-order DE

[KillerZ]

Homework Statement

$$y = c_{1}e^{x} + c_{2}e^{-x}$$ is a two-parameter family of solutions of the second-order DE $$y^{''} - y = 0$$. Find a solution of the second-order initial-value problem consisting of this differential equation and the given initial conditions.

Homework Equations

$$y(1) = 0$$
$$y^{'}(1) = e$$

$$y^{''} - y = 0$$
$$y = c_{1}e^{x} + c_{2}e^{-x}$$

The Attempt at a Solution

I am not sure if I found the solution correctly.

First derivative of the family of solutions:

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

Solving for $$c_{1}$$:

$$0 = c_{1}e^{1} + c_{2}e^{-1}$$
$$c_{1} = -\left(\frac{c_{2}}{e^{2}}\right)$$

Solving for $$c_{2}$$:

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

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

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

 

[Billy Bob]

Looks good. c_1 simplifies to 1/2.

 

[KillerZ]

Great thanks.