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.