# I can't solve this diff eqtn

1. Jun 30, 2011

### flyingpig

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

Find a particular soln

$$y'' -5y' +6y = te^t$$

The characteristic eqtn is $$r^2 - 5r + 6 = (r - 2)(r - 3)$$

r = 2, 3

According to my book

[PLAIN]http://img202.imageshack.us/img202/4539/unledxe.jpg [Broken]

So I thought

$$y = t(At + A_0)e^{2t} + t(Bt + B_0)e^{3t}$$

But not working...

what exatly is a simple root?

Last edited by a moderator: May 5, 2017
2. Jun 30, 2011

### micromass

Staff Emeritus
Hi flyingpig!

You seemed to use r=2 and r=3, which is incorrect.
What is the correct r? You only need to look at the right-hand side of

$$y^{\prime\prime}-5y^\prime+6y=te^t$$

3. Jun 30, 2011

### Staff: Mentor

To elaborate on what micromass said, r = 2 and r = 3 are roots of the characteristic equation, meaning that e2t and e3t are solutions of the homogeneous equation y'' - 5y' + 6y = 0. All solutions of this equation have the form yc = c1e2t + c2e3t, where yc is the complementary solution (the solution to the homogeneous problem).

What you're asked for is a particular solution of the equation y'' - 5y' + 6y = tet. This solution will involve et, but won't involve e2t or e3t.

4. Jun 30, 2011

### hunt_mat

I would try a particular solution of the form:
$$y_{p}=ate^{t}+be^{t}$$

5. Jul 1, 2011

### flyingpig

But that's what (14) says

6. Jul 1, 2011

### hunt_mat

and did you try it?

7. Jul 1, 2011

### flyingpig

$$y = t(At + A_0)e^{t}$$?

8. Jul 1, 2011

### ehild

No. It is not the same function hunt_mat suggested. Try that.

ehild

9. Jul 1, 2011

### Staff: Mentor

You're not reading the problem correctly, particularly item (i).
"s = 0 if r is not a root of the associated auxiliary equation."

10. Jul 2, 2011

### flyingpig

Yeah I don't understand what dose that mean? r is a root, i have two??

11. Jul 2, 2011

### micromass

Staff Emeritus
r has nothing to do with the characteristic equation. r is determined by the right-hand side of the equation, that is: $te^t$.
Thus t=1 here.

Now, to determine s, the characteristic equation does come into play. You'll need to check whether r is a root of the characteristic equation.

12. Jul 3, 2011

### flyingpig

That clears up ALOT.

Just one other thing. Is the table one of those things yo jsut have to remember?