Help with Laplace Transformations and 2nd order ODEs

[TFM]

Homework Statement

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}

Homework Equations

N/A

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

 

[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.

 

[TFM]

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.

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

TFM

 

[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.

 

[TFM]

Okay let's 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:

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

L(y(t))=y (in terms of s)

?

TFM

 

[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

 

[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

 

[sara_87]

you made a small mistake,
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)

 

[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

 

[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?

 

[TFM]

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?

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

 

[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?

 

[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

 

[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.

 

[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

 

[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?)

 

[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

 

[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)
but your steps are correct and so your Y would be:
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?

 

[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

 

[sara_87]

no, carefull... when i say inverse i don't 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)

 

[TFM]

The closest I can find is:

\frac{e^{-at}-e^{-bt}}{b - a} = \frac{1}{(p + a)(p + b)}

or

\frac{ae^{-at}-be^{-bt}}{b - a} = \frac{p}{(p + a)(p + b)}

?

TFM

 

[sara_87]

ok, let's first look at: 3/(s-1) since it's simpler:

L(e^t)= 1/(s-1) as we saw before
so to get 3/(s-1) this means the function is 3e^t
yes?

 

[TFM]

Yes, that makes sense and agrees with the similar one in the table.

TFM

 

[sara_87]

ok, so now what is the inverse of 2/(s-1)^2 ?
You have to use the first shift theorem which should be on the tables.

 

[TFM]

Would it not be:

2 \frac{1}{(p + a)(p + b)}

with a and b both minus 1 giving:


\frac{2}{(p - 1)(p - 1)}

?

TFM

 

[sara_87]

when you find the function, it should be in terms of t. eg:
inverse laplace of 3/(p-1) is 3e^t (from tables) (in terms of t's no p's; the p's are only in the laplace not the inerse of the laplace.
inverse laplace of 2/(p-1)^2 is...? find from tables, do you know the first shift theorem?

 

[TFM]

Its not on the table, and I can't seem to find it in my notes...

TFM

 

[sara_87]

Ok, the first shift theorem says that:
if you have a function that looks like: e^(at)F(t);
the laplace of that is the laplace of F(t) but instead of putting p, you put p-a
eg: L(e^3t(t))
we know that laplace of t is 1/p^2 (you know that from the tables right?)
and so here a=3 so the laplace of e^(3t)(t)=1/(p-3)^2
so, where ever you have p, you replace with p-a in this case, p-3
Do you agree??

so could you now find the inverse of 2/(s-1)^2
do have any ideas?

 

[TFM]

Thus bit is right?:

\frac{3}{(p-1)} : 3e^t

okay. so now I have to use first shift theorem on:

2/(p-1)^2

Would this not give:

2/((p - 3)-1)^2

I think that bit is wrong though?

Also, my table doesn't have t on its own, but I will accept that t goes to 1/p^2

?

TFM

 

[sara_87]

your table has t^n so just say that n is one to give laplace is 1/p^2

so, for 2/(p-1)^2
why did you do this: 2/((p-3)-1)^2 ?
You have to find a function in t such that when you find the laplace of it, you would get 2/(p-1)^2. So, don't apply the first shift theorem on it, you have to unapply the first shift theorem...does that make sense?

Hint: what is the laplace of 2t ? (using tables with n=1)

 

[TFM]

Missed that one...

so 2t:

\frac{2}{p}^2

?

TFM

 

[TFM]

I've written it slightly upside down!

should've been:

\frac{2}{p^2}

Sorry...

TFM

 

[sara_87]

laplace of 2t is 2(1/p^2) = 2/p^2
agree??

so, then you use the first shift theorem since we don't want 2/p^2; we want 2/(p-1)^2
so do you what to do then??

 

[TFM]

I agree with that. So I am assuming that now I need to use the First Shift Theorem:

\frac{2}{p^2}

using the example of:

e^{(3t)}(t)=1/(p-3)^2

\frac{2}{p^2} = 2t - 2

Does this look right?

TFM

 

[sara_87]

Laplace of e^3t(t)=1/(p-3)^2
so Laplace of ------ = 2/(p-1)^2
we know that laplace of 2t=2/p^2
so that means laplace of 2t(e^t)=2/(p-1)^2
does this make sense to you?? (think about it). you are finding the opposite of the laplace...called the inverse.

 

[TFM]

Laplace of e^3t(t)=1/(p-3)^2
so Laplace of ------ = 2/(p-1)^2
we know that laplace of 2t=2/p^2
so that means laplace of 2t(e^t)=2/(p-1)^2
does this make sense to you?? (think about it). you are finding the opposite of the laplace...called the inverse.

That does indeed make sense, so now would you put it all together (ie the e^3t and the 2t(e^t))?


TFM

 

[sara_87]

yes... :)
now your function is:
y(t)=2te^t+3e^t

So, hopefully now, you could do the other questions you posted in the first post. but remember:
L(y(t))=Y
L(y'(t))=sL(y(t)) - y(0)
L(y''(y))=sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0)
etc.

 

[TFM]

Excellent.

Maybe I could go through one myself on here now?

TFM

 

[sara_87]

yep
deffinetely

 

[TFM]

Okay so (b)

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

remembering:

L(y(t))=Y
L(y'(t))=sL(y(t)) - y(0)
L(y''(t))=sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0)

L(4y(t))=4Y

L(y'(t))=sL(4y(t)) - y(0)

L(y''(t))=sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0)

So Far, so good?

TFM

 

[sara_87]

so far so good!

 

[TFM]

Excellent. Now:

L(4y(t))=4Y

L('(t))=sL(4y(t)) - y(0)

L(y''(t))= s^2(L(y(t))-sy(0)-y'(0)

Insert values for y(0) and y'(0):

L('(t))=sL(4y(t)) - 0 = sL(4y(t))

L(y''(t))= s^2(L(y(t))-s0 - 4 = s^2(L(y(t)) - 4

I think this is where it starts going down hill slightly:

L('(t)) = sL(4y(t))

Insert y(t):

L('(t)) = sL(4Y)

L(y''(t)) = s^2(L(Y) - 4

Does this still look right?

TFM

 

[sara_87]

ok, i think you get the idea but you are getting mixed it up a bit with y and Y. this is what you should get:
L(y(t))=Y (which is what you got) (dont put the 4 in yet you will see why later)
L(y'(t))=sL(y(t))-y(0) = sY-0 = sY
L(y''(t))=sL(y'(t))-y'(0)=s(sY)-4 = s^2(Y)-4

If you did L(4y(t)), L(y'(t))=s(L(4y(t)))-y(0) which is not true since L(y'(t))=sL(y(t))
Does that make sense?

 

[TFM]

Okay, so:

L(y(t))=Y

L(y'(t))=sL(y(t))-y(0) = sY-0 = sY

L(y''(t))=sL(y'(t))-y'(0)=s(sY)-4 = s^2(Y)-4

Now I assume we use the Laplace transformations from the Table?

TFM

 

[sara_87]

Now, you substitute each one back into your original equation:
y''+4y'+4y=L(e^-2t)
(and don't forget to find the laplace of e^-2t using the rule we used before)

 

[TFM]

Okay so:

L(e^{-2t})

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

This makes alpha 2

e^{-2 t} = \frac{1}{p + 2}

So now we insert these values into original equation:

y''+4y'+4y=L(e^{-2t})

giving:

(S^2(Y) - 4) + 4(sY) + 4Y = \frac{1}{p + 2}

I think the p is the same a s:

(s^2(Y) - 4) + 4(sY) + 4Y = \frac{1}{s + 2}

Does this look okay?

TFM

 

[sara_87]

yep that's perfectly fine; now, you have:
Ys^2-4+4sY+4Y=1/(s+2)

so make Y the subject and find the inverse.

 

[TFM]

okay so:

s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2}

put the lonsome 4 on the otherside:

s^2(Y) + 4(sY) + 4Y = \frac{1}{s + 2} + 4

Factorise:

Y(s^2 + 4(s) + 4) = \frac{1}{s + 2} + 4

divide through:

Y = \frac{\frac{1}{s + 2} + 4}{(s^2 + 4(s) + 4)}

This looks rather complicated...

TFM

 

[sara_87]

notice that the denominator can be factorized to: (s+2)^2

so, now you have:
Y=1/(s+2)^3 + 4/(s+2)^2

now find the inverse laplace of that (hint: you will need to use the first shift theorem as we did before)

 

[TFM]

I missed that one...

So:

Y = \frac{\frac{1}{s + 2} + 4}{(s^2 + 4(s) + 4)} \equiv \frac{\frac{1}{s + 2} + 4}{(s + 2)^2}

I can't see any similar functions in the table, however?

TFM