# Second Diff Eq question of the day

1. Feb 8, 2009

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

10. If y(t) = -te^(3t) is a solution to y'' + py' + qy = 0, then what are p and q?

2. Relevant equations

3. The attempt at a solution

so it's equal to r^2 + pr +q = 0, but then how can i find the two numbers?

thanks

2. Feb 8, 2009

### rootX

You don't need to do any algebra. Remember there are some formulas for solving different equations? This one looks like e^at (At + B) form. Just look over them.

3. Feb 8, 2009

### Staff: Mentor

If your solutions were y1 = e^(3t) and y2 = e^(4t), your characteristic equation would be r^2 - 7r + 12 = 0, or (r - 3)(r - 4) = 0, right?

Under what circumstances related to the work above will you get a solution y = te^(3t)?

4. Feb 9, 2009

### HallsofIvy

Staff Emeritus
You don't need to look at the characteristic equation at all. If $y= -te^{3t}$ then $y'= -e^{3t}- 3te^{3t}= -(1+ 3t)e^{3t}$ and y"= $-3e^{3t}- 3(1+ 3t)e^{3t}$$= -(6+ 9t)e^{3t}$. Putting those into the equation we have $-(6+ 9t)e^{3t}- p(1+ 3t)e^{3t}- qte^{3t}= ((-6- p)+(-9-3p-q)t)e^{3t}= 0$ for all t. Since $e^{3t}$ is never 0, we must have (-6- p)+ (-9-3p-q)t= 0 for all t which means we must have -6-p= 0 and -9- 3p- q= 0.

Of course, it is also true that eat will be a solution if and only if a is a root of the characteristic equation and that teat will be a solution if and only if a is a double root of the characteristic equation, which means that the characteristic equation must reduce to (x-a)2= 0.

Last edited: Feb 9, 2009