Can someone verify my solution?

[CentreShifter]

I think there may be a typo in the book, I'm pretty sure I'm doing this correctly.

Use the Laplace transform to solve the IVP: y"-6y'+9y=t; y(0)=0, y'(0)=0

My solution is e$$^{3t}$$(1/9*t - 2/27) + 1/9*t + 2/27.

Can someone quickly solve it again for me?

 

[tiny-tim]

I think there may be a typo in the book, I'm pretty sure I'm doing this correctly.

Use the Laplace transform to solve the IVP: y"-6y'+9y; y(0)=0, y'(0)=0

My solution is e$$^{3t}$$(1/9*t - 2/27) + 1/9*t + 2/27.

Can someone quickly solve it again for me?

Hi CentreShifter!

y"-6y'+9y = what?

 

[arildno]

I think there may be a typo in the book, I'm pretty sure I'm doing this correctly.

Use the Laplace transform to solve the IVP: y"-6y'+9y; y(0)=0, y'(0)=0

My solution is e$$^{3t}$$(1/9*t - 2/27) + 1/9*t + 2/27.

Can someone quickly solve it again for me?

Well, what is your differential EQUATION??

Is it: y"-6y'+9y=0 ?

First, we identify that [itex]Ae^{3t}[/tex] is, indeed, a double root-solution of the IVP

In order to find a second solution, we try with:
$$Bte^{3t}$$
Inserting this trial solution into our equation yields:
$$(6Be^{3t}+9Bte^{3t})-6(Be^{3t}+3Bte^{3t})+9Bte^{3t}=0$$
Note that simplification of the left-hand side yields:
$$0=0$$

This is precisely what you should have, since you now have two arbitrary parameters, A og B, by which you may adjust your general solution, $$y=Ae^{3t}+Bte^{3t}$$, to the initial conditions.

(Note that this will yield you y=0 as your solution, do you now realize WHY you must state precisely what your diff. eq. actually was?

 

[CentreShifter]

You are both absolutely correct. I was the end of my study session, there should definitely be an equation there. I'll be posting it as soon as I can get to the book.

@arildno - I know this doesn't help right now, but the problem is to be solved using Laplace transforms, not undetermined coefficients (although I suppose t doesn't really matter as long as the solution is correct).

Edit: I have fixed the equation in the first post. It's now correct.