# Homework Help: Help with Laplace Transformations and 2nd order ODEs

1. Nov 23, 2008

### TFM

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

Solve the following problems using Laplace Transforms:

$$y' - y = 2e^t, y_0 = 3$$

$$y'' + 4y' + 4y = e^{-2t}, y_0 = 0, y_0' = 4$$

$$y'' + y = sin(t), y_0 = 1, y_0' = 0$$

$$y'' + y = sin(t), y_0 = 1, y_0' = -\frac{1}{2}$$

2. Relevant equations

N/A

3. The attempt at a solution

I don't seem to understand how to use Laplace Transformation to solve ODE's! If anyone can help, I will be most grateful.

TFM

2. Nov 23, 2008

### sara_87

lets look at the first one: let L(F(t)) denote the laplace of F(t)
in general, you say that:
L(y'(t))-L(y(t))=L(2e^t) call this (1)

L(y(t))=y (in terms of s)
L(y'(t))=sL(y(t) - y(0) (This is one of the formulas from the table of laplace transforms)
=sy-3
so, substituting these back into (1): sy-3-y=2/(s-1) (also from the tables)

make y the subject, then you should get an expression for y in terms of s.
Then to find y(t) (which is what you are looking for), you must find the inverse of y.

I hope all is clear.

3. Nov 23, 2008

### TFM

Where did the 'in terms of S' come from?

TFM

4. Nov 23, 2008

### Pere Callahan

It might be confusing to use the same label y for the function und its Laplace transform. Better come up with a new letter like Y

You then have to solve (according to Sara)
$$sY(s)-3-Y(s)=\frac{2}{s-1}$$
for Y(s) and find the inverse Laplace transform of this function.

5. Nov 23, 2008

### TFM

Okay lets see if this makes sense:

you have the orginal equation, F(t)

$$y' - y = 2e^t$$

then you make L(f(t)):

$$L(y'(t)) - L(y(t)) = L(2e^t)$$ ---(1)

I still can't see how you have gone from here to here:

???

TFM

6. Nov 23, 2008

### sara_87

that is not from one step to another.
L(y'(t))-L(y(t))=L(2e^t) this is something
and
L(y(t))=Y is something else.
In other words, LET L(y(t))=Y
so therefore, L(y'(t))=sY-3

7. Nov 23, 2008

### TFM

Let's see then...:

$$L(F(t): L(y'(t)) - L(y(t)) = L(2e^t)$$ ---(1)

and also:

$$L(y(t)) = Y$$

so inserting into (1)

$$L(F(t): L(y'(t)) - Y = L(2e^t)$$

$$L(y'(t)) = Y + L(2e^t)$$

I see now where the 3 comes from, $$y_0 = 3$$

$$L(y'(t)) = Y - y_0$$

$$L(y'(t)) = Y - 3$$

substitute back into (1):

$$Y - 3 - L(y(t)) = L(2e^t)$$

does this look right so far?

TFM

8. Nov 23, 2008

### sara_87

L(y'(t))=sY-3
because you must use the formula:
L(F'(t))=sL(F(t))-F(0)
Im not sure but I think you are expected to know this general formula.
so this gives:
sY-3-L(y(t))=L(2e^t)

9. Nov 23, 2008

### TFM

Yes, the formula:

$$L(F(t)) = sL(F(t)) - F(0)$$

looks useful. I assume the $$sL(F(t))$$ is inequivalent to: $$s*L(y(t)) = Y$$

???

so:

$$L(y'(t)) = sY - 3$$

putting back into (1)

$$L(y'(t)) - L(y(t)) = L(2e^t)$$

$$sY - 3 - sY - 3 = L(2e^t)$$

does this look right?

TFM

10. Nov 23, 2008

### sara_87

The star means convolution (in Laplace language) so it is inequivalent
You have:
L(y'(t))-L(y(t))=L(2e^t) this is (1)
we said let L(y(t))=Y
and then we know that L(y'(t))=sY-3

so putting into (1) gives:
sY-3-Y=L(2e^t)

Why are you putting sY-3 twice?

11. Nov 23, 2008

### TFM

I meant the star as in times, Sorry.

I am not sure, how am I putting in sY-3 twice? sY-3-Y=L(2e^t) only has sY - 3 in once?

TFM

12. Nov 23, 2008

### sara_87

exactly so again you have to remember:
L(y'(t))-L(y(t))=L(2e^t) this is (1)
(A) we said let L(y(t))=Y
(B) and then we know that L(y'(t))=sY-3
so putting into (1) gives:
(C) sY-3-Y=L(2e^t)
These are the three main steps that should give you the equation of Y in terms of s.
So, what would you do next?

13. Nov 23, 2008

### TFM

Well, judging from your first post, rearrange into Y?

so this would be:

$$sY-3-Y=L(2e^t)$$

put 3 to one side:

$$sY-Y=L(2e^t) + 3$$

factorise out Y:

$$Y(s - 1) = L(2e^t) + 3$$

divide by (s - 1):

$$Y = \frac{L(2e^t) + 3}{s - 1}$$

Does this look right?

???

TFM

14. Nov 23, 2008

### sara_87

no, before you 'put to one side' on your second step, you have to find the laplace of
2e^t
sY-3-Y=L(2e^t)
so,
sY-3-Y=2/(s-1)
(do you know that L(2e^t)=2/(s-1) ?)
now, you make Y the subject.

15. Nov 23, 2008

### TFM

I have a table with Laplace conversions, and it gives:

$$e^{-\alpha t} = \frac{1}{p + \alpha}$$

so

$$2e^{t} = \frac{2}{p + t}$$

I'm assuming p and s are equivilant

so now:

$$sY-3-Y = L(2e^t)$$

$$sY-3-Y = \frac{2}{s + t}$$

now take over the 3:

$$sY-Y = \frac{2}{s + t} + 3$$

factorise Y:

$$Y(s - 1) = \frac{2}{s + t} + 3$$

thus giving Y

$$Y = \frac{\frac{2}{s + t} + 3}{s - 1}$$

Does this look better?

TFM

16. Nov 23, 2008

### sara_87

e^(-at)=1/(s-a)
so: (the right hand side is the laplace so it has nothing to do with t)
2e^(t)=1/(s+1) (since in our case, -a=1)

Now, make Y the subject from:

sY-3-Y=1/(s+1) (do you agree?)

17. Nov 23, 2008

### TFM

I feel slightly sill, I meant to do that, but didn't

So:

$$sY-3-Y = \frac{2}{s + 1}$$

now:

$$sY-Y = \frac{2}{s + 1} + 3$$

$$Y(s - 1) = \frac{2}{s + 1} + 3$$

$$Y = \frac{\frac{2}{s + 1} + 3}{s - 1}$$

Is this right?

TFM

18. Nov 23, 2008

### sara_87

Im sorry, i misread; the L(2e^t) means that our a=1 so that means L(2e^t)=2/(s-1)
not 2/(s+1)
sY-3-Y=2/(s-1)

Y(s-1)=2/(s-1) +3
so Y=2/(s-1)^2 + 3/(s-1) (by dividing both sides by s-1)
is this clear?
so, what would you do next?

19. Nov 23, 2008

### TFM

so:

$$Y = \frac{\frac{2}{s - 1} + 3}{s - 1}$$

is the same as:

$$Y = \frac{2}{(s - 1)^2}+\frac{3}{s - 1}$$

so now, from your first post again, I need to find the inverse:

thus:

$$\frac{1}{Y} = \frac{(s - 1)^2}{2}+\frac{s - 1}{3}$$

???

TFM

20. Nov 23, 2008

### sara_87

no, carefull.... when i say inverse i dont mean 1/Y;
Y=2/(s-1)^2 + 3/(s-1) call this (2) (This is you Laplace of the function you are looking for)

So you want to find the function in terms of t that would give you (2) when you find its laplace. so you want to find a function of t.
so, Y(t)= something that when you find its laplace you would get (2)

so, first, what is the function in terms of t that when you find its laplace you would get 2/(s-1)^2 ? (you would need to use your tables)