# Homework Help: Second Order Differential Equations

1. Jun 2, 2012

### roam

1. The problem statement, all variables and given/known data

Solve the following initial value problem:

$\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = t$

$y(0) = 1, \ y'(0) = 0$

The correct answer must be: $y(t)=3e^{-t} + 2 t e^{-t} + t -2$

3. 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:

$\frac{d^2y}{dt^2} + 2 \frac{dy}{dt}+y = 0$

$\frac{d^2 e^{st}}{dt^2} + 2 \frac{de^{st}}{dt}+e^{st} = (s^2 + 2s +1) e^{st} =0$

$(s+1)(s+1) \implies s=-1$

So we have a repeated eigenvalue. Solution to the homogeneous equation is

$y_h= C_1 e^{-t}$

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.

2. Jun 2, 2012

### Staff: Mentor

No, this is only one solution, and you need two. In cases like this where there is a repeated root of the characteristic equation, you get another solution by multiplying by t.

So two solutions of the homogeneous equation are
$\{e^{-t}, te^{-t}\}$

The solution to the homogeneous equation would be all linear combinations of the above; $y_h= C_1 e^{-t} + C_2t e^{-t}$
A better particular solution to the nonhomogeneous problem would be yp = A + Bt

3. Jun 3, 2012

### roam

Oh I see, that makes perfect sense to me know. Thank you so much for your help, I really appreciate it.