# Laplace transform of y''(t')

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1. Nov 4, 2015

### Linder88

• Member warned about posting without the HW template
The ordinary differential equation, with initial values,shall be solved using Laplace transform. The ODE looks like this

y''(t')+2y''(t)-2y(t)=0

And the initial conditions are

y(0)=y'(0)=0, y''(0)=0

The problem is with the first term in the ODE, how do I transform taking the derivative of t in consideration? I would be happy for an answer!

Last edited by a moderator: Nov 4, 2015
2. Nov 4, 2015

### Ray Vickson

There are standard results linking the L.T.s of y'(t) and y''(t) to the L.T. of y(t). You can (and should) look them up in any table of Laplace Transforms. (PF Rules require that you show some effort before receiving help.)

Last edited by a moderator: Nov 4, 2015
3. Nov 4, 2015

### Linder88

I think you misunderstood the question, the problem is not to make Laplace transforms of y(t), y'(t) and y''(t) but of y''(t'). There's a certain difference because taking the next step in the solution is dependent of this question.

4. Nov 4, 2015

### Staff: Mentor

1. The homework template is required. In future posts, please do not delete it.
2. In your diff. equation, the dependent variables are y and y''; the independent variable is t. I would guess that the expression t' means $\frac {d}{dt}(t) = 1$. This would mean that your DE is y''(1) + 2y''(t) - 2y(t) = 0, or equivalently, y'' - y = -(1/2)y''(1).
The only other thing I can think of is that t' represents some transformation of t.
3. Your begin and end commands used the wrong slash -- it should have been \, not /.

5. Nov 4, 2015

### Ray Vickson

I thought your question contained a "typo", and that you had intended y'''(t), because I could not believe anybody would write y''(t') if t is the independent variable. Of course if t = f(w) is actually some known(?) function of another variable w---as suggested by Mark---then the question would make sense if t' means f'(w). On the other hand, if t' really does mean dt/dt = 1, then y''(t') is just some constant (i.e., does not vary with t), so its L.T. is just the L.T. of a constant.

6. Nov 4, 2015

### Linder88

Thank you so much for your answer, it all makes sense! All three points :)