Help with Laplace Transformations and 2nd order ODEs

[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

 

[sara_87]

split this into two fractions:
Y=1/(s+2)^3 + 4/(s+2)^2

now as we did before, the second fraction is the laplace transform of: 4te^(-2t)
because if you find the laplace of 4t, this will give you 4/s^2 but we don't want s at the bottom, we want (s+2) so that's why we use the first shift theorem and times the 4t by e^(-2t). the sfirst shift theorem must be in your tables or at least in your notes.

So, i did the second fraction for you (the easy one :) ) and so now you do the first fraction. what is the inverse laplace of 1/(s+2)^3 ? and remember,
the laplace of t^n= (n!)/(s^(n+1))

 

[TFM]

so split it up:

$$\frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$

$$\frac{4}{(s + 2)^2} = 4te^{-2t}$$

okay so for:

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

we need to use:

$$t^n = \frac{n!}{s^{n + 1}}$$

so, would this mean:

$$\frac{3!}{s^3} = t^2$$

we have

$$\frac{1}{s^3}$$

Is this relevant, or am I going the wrong way for this one/

TFM

 

[sara_87]

for 1/(s+2)^3:
laplace of t^2 is 2!/(s^3) so it is 2/s^3
i don't know why you put 1/s^3.
but you have the right idea.
so if the laplace of t^2 is 2!/(s^3), what is the inverse of 1/(s+2)^3 ?

 

[TFM]

So:

$$t^2 = \frac{2!}{s^3}$$

we have $$\frac{1}{(s + 2)^2}$$

so using the Shift, would that be something like:

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

bvut I think even it is slightly wrong at least because it doesn't deal with the fact that the top value isn't 2!

TFM

 

[sara_87]

For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right?
but, we don't want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?

 

[TFM]

First Shft Theorem

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??

$$t^2 = 2/s^3$$

$$\frac{1}{2}t^2 = \frac{1}{s^3}$$

using FST

$$\frac{1}{2}t^2 = \frac{1}{b^3}$$, where b = s + 1

would this make it:

\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]

?

TFM

 

[sara_87]

I don't see any e's in your answer (there should be).

now as we did before, the second fraction is the laplace transform of: 4te^(-2t)
because if you find the laplace of 4t, this will give you 4/s^2 but we don't want s at the bottom, we want (s+2) so that's why we use the first shift theorem and times the 4t by e^(-2t).

now, can you do the same for the 1/(s+1)^3

 

[TFM]

Okay so:

4t will give you 4/s^2,but we don't want s at the bottom, we want (s+2) so that's why we use the first shift theorem and times the 4t by e^(-2t).

so would this mean:

$$t^2 would give: \frac{2}{s^3}$$

thus would:

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

?

TFM

 

[sara_87]

very good!

 

[TFM]

Great

So now:

$$Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$

this goes to:

$$Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$

So now we put this value of Y back into:

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

?

TFM