- #1
roam
- 1,271
- 12
Homework Statement
Solve the following initial value problem:
[itex]\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = t[/itex]
[itex]y(0) = 1, \ y'(0) = 0[/itex]
The correct answer must be: [itex]y(t)=3e^{-t} + 2 t e^{-t} + t -2[/itex]
The Attempt at a Solution
Where did they get the term 2te-t from?
Here's how I've done it so far:
We can find the general solution of the corresponding homogeneous system:
[itex]\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = 0[/itex]
[itex]\frac{d^2 e^{st}}{dt^2} + 2 \frac{de^{st}}{dt}+e^{st} = (s^2 + 2s +1) e^{st} =0[/itex]
[itex](s+1)(s+1) \implies s=-1[/itex]
So we have a repeated eigenvalue. Solution to the homogeneous equation is
[itex]y_h= C_1 e^{-t}[/itex]
Now we must find a particular solution, I chose yp=t-2, since we it is a solution of the nonhomogeneous equation. Therefore the solution is
y = yh + yp = C1 e-t + t - 2
Now I can use the given initial values to work out C1:
y' = -C1e-t + 1
y(0) = C1 - 2 = 1 ⇒ C1 = 3
y'(0) = -C1+1 = 0
I get y = 3e-t + t - 2, but there must be another coefficient so we can solve two equations in two unknowns. But how do we get the term 2te-t? Even if I wrote the repeated eigenvalue twice C1e-t + C2e-t, I still don't get the term 2te-t. So what's the problem?
Any help is greatly appreciated.