# Differential equation with laplace transform

1. Oct 22, 2009

### oddiseas

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

Solve:

ty''+2y'+y=tJ₂(2√t)

with y(0)=y'(0)=0

2. Relevant equations

3. The attempt at a solution

Applying the laplace transform i get:

L(y)=Y
L(2y')=2sY
L(ty'')=-2sY-s^2Y

Putting this together:
-2sY+2sY+Y-s^2(dy/ds)=[e^-(1/s)]/s^3

Y'-(1/s^2)Y=[e^-(1/s)]/s^3

Which i can solve because it is first ordre non homogeneous> solving i get:

Y(s)=[e^-(1/s)]/(4s^4 ) +Ce^-(1/s)

I am stuck on how to aplly the initial value theorm. That is the limit as t approches zero of f(t) is equal to the limit as f(s)*s approaches infinity. Apllying this rule i get:

[e^-(1/s)]/(4s^3 ) +sCe^-(1/s)

and as s approches infinity this approaches infinity not zero. So what do i do now? to get the right solution?

2. Oct 22, 2009

### oddiseas

L(tY'')=-s^2(dY/ds)-2sY

I did not make a mistake in the actual calculation, this was a typing error.

3. Oct 23, 2009

### HallsofIvy

Staff Emeritus
This makes no sense. You should not have a "Y' ". The whole point of the "Laplace transform" is that it converts a differential equation for y to an algebraic equation for Y.

There should be no "dy/ds" in the equation.

Using what you have written above, $2sY- s^2Y+ 2sY+ Y= e^{-1/s}/s^3$

Solve that for Y.

4. Oct 23, 2009

### oddiseas

The inverse transorm of a function multiplied by "t" to any power n, is the nth derivative of the transform with respect to s, without the "t" factor:
ie -1^n d/ds
Thus L(Y't)=-d/ds(sY) etc.
this allows us to transpose a differential equation with "variable coefiicients" with respect to t. If you dont know what youre talking about, dont reply to my posts thanks>

Last edited by a moderator: Oct 23, 2009
5. Oct 23, 2009

### Staff: Mentor

I can vouch for the fact that HallsOfIvy does know what he's talking about.
Not quite. The Laplace transform (not the inverse (Laplace) transform) of tny(t), or L{tny(t)}, is (-1)nY(n)(s).

In your differential equation, you need the Laplace transform of ty''(t). The formulas above doesn't cover this. I was unable to find the Laplace transform of ty''(t) in 6 different online tables, so I can't confirm what you have or say that it's wrong. If I continued to be unsuccessful in finding a formula for the Laplace transform of this function, I would try to get it by using the definition, namely
$$L\{ty''(t)\} = \int_{0}^{\infty} t y''(t) e^{-st} dt$$
No, L{y'(t)} = sY(s) - y(0), so you're not even warm here. You are confusing two different formulas L{ty(t)} and L{y'(t)}.

Last edited by a moderator: Oct 23, 2009
6. Oct 23, 2009

### oddiseas

Are you telling me that "Both" my mathematics professors who confirmed this is the right method to use, and showed us the steps in solving these differential equations are wrong and you are right? Check your maths books for variable cooeficient, and by the way, just because you "vouch" for someone does that mean that they are correct. Maybe you should investigate the concept thorouly for yourself and then make an informed statement.

Last edited by a moderator: Oct 23, 2009
7. Oct 24, 2009

### Staff: Mentor

Please read what I wrote in my previous post. I said nothing about your method being an incorrect method. What I did do was point out two errors in what you wrote, one minor and one where you were way off. The formulas I provided can be found in any halfway complete table of Laplace transforms. If there is anything in my post that is incorrect, I would like to know about it.

To be honest, though, your attitude surprises me. It is not often that students come here for help, and insult those who are trying to help them.
Mark

8. Oct 24, 2009

### lurflurf

ty''(t) is not normally listed in tables, one is expected to combine the results of t y and y''

as for
L{t y'(t)}=-[L{y'(t)}]'=-[sY(s)-y(0)]'=-[sY(s)]'=-Y(s)-s Y'(s)

9. Oct 24, 2009

### lurflurf

Just because someone makes a mistake does not mean you should be rude.