Help with Laplace Transformations and 2nd order ODEs

[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

 

[sara_87]

y(t)=(1/2)t^2(e^-t) + 4t(e^-2t)
This is the answer. don't substitute anything into anything.

 

[TFM]

Excellent.

So

y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}

Tomorrow, I'll start (c), if that's okay?

TFM

 

[sara_87]

yep that's fine

 

[TFM]

Okay, so:

(c):

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

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)


there is no y',


L(y(t)) = Y
L(y'(t)) = sL(y(t)) - 1
L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0)


and for the sin(t)

sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2}

since alpha = 1:

sin(t) = \frac{1}{p^2 + 1}

Is this okay, I am not quite sure because the lack of y'

TFM

 

[sara_87]

That is absolutely right.
what next?

 

[TFM]

Okay so:

L(y(t)) = Y

L(y'(t)) = sL(y(t)) - 1

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

So now I substitute these into the original equation:

y'' + y = sin(t)

s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}

Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone?

TFM

 

[sara_87]

That's fine
substitute y(0) and also substitue the value L(y(t))

 

[TFM]

Okay, so

s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}

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

Ys^2 + Y - 1 = \frac{1}{p^2 + 1}

Now I need to make the Y the subject:

Ys^2 + Y = \frac{1}{p^2 + 1} + 1

factorise out

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

divide through:

Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1}

Now I need to find the inverse.

Firstly, split it up into two fractions:

Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1}

Does this look okay?

TFM

 

[sara_87]

in ur second line, you have: s^y-1+y=1/(s^2+1)
the -1 on the left hand side should be -s (since you have:L(y'')=s(sy-1)
and remember: s is p so don't put both...choose one of them :)

 

[TFM]

Okay so:

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

Ys^2 + sY - 1 = \frac{1}{s^2 + 1}

Does this look okay now?

TFM

 

[sara_87]

you should have:
Ys^2 - s + Y=1/(s^2+1)

i think you made a mistake while substituting.

 

[TFM]

Okay, back a bit originally:

s^2(L(y(t))-sy(0) + Y = \frac{1}{s^2 + 1}

y(0) = 1

L(y(t)) = Y

put these in:

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

goes to:

Ys^2 - 1 + Y = \frac{1}{s^2 + 1}

I still have a minus one where there should be a minus s.?

TFM

 

[sara_87]

y''+y'=sint

L(y(t))=Y
L(y'(t)=sL(y(t))-y(0)=sY-1
L(y''(t))=sL(y'(t))-y'(0)=s(sy-1)-0
=s^2Y-s
substitute this into the equation: y''+y'=sint
(s^2)Y-s+Y=1/(s^2+1)

 

[TFM]

y''+y'=sint

L(y(t))=Y
L(y'(t)=sL(y(t))-y(0)=sY-1
L(y''(t))=sL(y'(t))-y'(0)=s(sy-1)-0
=s^2Y-s
substitute this into the equation: y''+y'=sint
(s^2)Y-s+Y=1/(s^2+1)

Isn't the equation:

y'' + y = sin(t)

though, not

y'' + y' = sin(t)

?

TFM

 

[sara_87]

yep that's what i meant.

substitute this into the equation: y''+y=sint
(s^2)Y-s+Y=1/(s^2+1)

 

[TFM]

Okay so:

<br /> y&#039;&#039;+ y = sin(t) <br />

L(y(t))=Y

L(y''(t))= s(sy-1)-0

insert into equation:

s(sy - 1)+ Y = sin(t)

multiply out:

s^2y - s+ Y = sin(t)

sin(t)= 1/(s^2+1)

s^2Y - s + Y = \frac{1}{s^2+1}

Is this okay? If so I now have to rearrange to find Y

s^2Y + Y = \frac{1}{s^2+1} + S

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

giving:

Y = \frac{\frac{1}{s^2+1} + S}{s^2 + 1}

split the fraction:

Y = \frac{\frac{1}{s^2+1}}{s^2 + 1} + \frac{S}{s^2 + 1}

Does this look better?

TFM

 

[sara_87]

much better :)

so now you have:
Y=1/[((s^2)+1)^2] + s/((s^2)+1)

now, the inverse laplace of s/(s^2 + 1) should be in you tables (can you find it?) and so, what are you going to do about the first fraction 1/[((s^2)+1)^2] ?

 

[TFM]

the inverse of:

\frac{S}{s^2 + 1} is cos(t)

for:

\frac{1}{((s^2)+1)^2},

I assume the first Shift Theroem is required?

TFM

 

[sara_87]

no first shift theorem...why do you assume that??

1/[((s^2)+1)^2] can be split into 1/[(s^2)+1] * 1/[(s^2)+1]
convolution says that if you have the product of two functions, F(s)G(s), the inverse laplace can be evaluated by doing: first find the inverse laplace of F(s) and G(s) (call them F(t) and G(t) respectively) then carry out:
integral (with limits 0 to t) of (F(u)G(t-u)) du
the value of the integral gives you the inverse of the laplce F(s)G(s).

so in you case, you have: F(s)=1/(s^2+1) and G(s)=1/(s^2+1)
what's the inverse of both of them (surely they give the same inverse) ?

 

[TFM]

so we have:

F(s)=1/(s^2+1) and G(s)=1/(s^2+1)

the inverse of both is:

sin(t)

All right so far?

TFM

 

[sara_87]

yep that's right, so now evaluate the integral with limits 0 to t of:
sin(u)sin(t-u) du

you might wana remember the formula:
sin(A)sin(B)=1/2(cos(A-B)-cos(A+B))
and in this case, A=u and B=t-u
you have been taught convolution...right?

 

[TFM]

Okay so:

sin(A)sin(B)=1/2(cos(A-B)-cos(A+B))

A=u and B=t-u

sin(u)sin(t - u)=1/2(cos(u-(t - u))-cos(u+(t - u)))

this would go to:

1/2(cos(2u-t)-cos(t))

Integrate from 0 to t

\int^t_0 1/2(cos(2u-t)-cos(t)) du

Take the half outside:

\frac{1}{2}\int^t_0 cos(2u-t)-cos(t) du

integrating becomes:

\frac{1}{2} \left[ sin(2u - t) - sin(t)\right]^t_0

okay so far?

TFM

 

[sara_87]

whats the integral of: cos(t) du ?

 

[TFM]

Hmm,

wells, cos(t) has no values of u, so can be assumed to be a constant, so would it be cos(t)*u ?

TFM

 

[sara_87]

yep...and what is the integral of cos(2u-t)du ??

 

[TFM]

The integral of cos(u) is sin(u)
The integral of cos(2u) is 1/2sin(2u)

So would that make the integral of:

cos(2u - t) = 1/2sin(2u - t) ?

TFM

 

[sara_87]

good
so evaluate with limits t and 0 and don't forget the 1/2 outside the bracket.

 

[TFM]

Okay so now:

\frac{1}{2} \left[ 1/2sin(2u - t) - cos(t)u \right]^t_0

so this goes to:

\frac{1}{2} \left[ (1/2sin(2t - t) - cos(t)t) - (1/2sin(2(0) - t) - cos(t)0) \right]

and:

\frac{1}{2} \left[ (1/2sin(t) - cos(t^2) - (1/2sin(-t)\right]

Does this look okay now?

TFM

 

[sara_87]

the second to last step is:
1/2(1/2sin(2t-t) - tcos(t) - 1/2(sin(-t)) - 0cost)

the last step is not ok... remember: sint is odd function so sin(-t)=-sin(t)

so...how would you write the last step?

 

[TFM]

okay so:

\frac{1}{2} \left[ (1/2sin(2t - t) - tcos(t)) - (1/2sin(2(0) - t) - cos(t)0) \right]

\frac{1}{2} \left[ (1/2sin(2t - t) - tcos(t)) - (1/2sin(-t)) \right]

this is the same as:

\frac{1}{2} \left[ (1/2sin(2t - t) - tcos(t)) + (1/2sin(t)) \right]

Is this better?

TFM

 

[sara_87]

yep but it can be simplified more (the first term?? and expand the bracket)

 

[TFM]

How did I miss it that time? I did it before. So:

\frac{1}{2} \left[ (1/2sin(2t - t) - tcos(t)) + (1/2sin(t)) \right] [\tex]<br /> <br /> Can be simplified to:<br /> <br /> \frac{1}{2} \left[ (1/2sin(t) - tcos(t)) + (1/2sin(t)) \right] [\tex]&lt;br /&gt; &lt;br /&gt; Which can go to:&lt;br /&gt; &lt;br /&gt; \frac{1}{2} \left[ sin(t) - tcos(t) \right] [\tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Better?&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; TFM

 

[sara_87]

it's 1/2(-sint-tcost) because in the second to last step, 1/2sint should be -1/2sin(t) (odd function sine).

 

[TFM]

Okay so it should be:

\frac{1}{2} \left[ (-1/2sin(t) - tcos(t)) - (1/2sin(t)) \right]

Which goes to:

\frac{1}{2} \left[-sin(t)- tcos(t))\right]

So this is correct now?

TFM

 

[sara_87]

yes, now remember:
Y=1/(s^2+1)^2 + s/(s^2+1)

the 2nd fraction is the laplace of cost
the 1st fraction is the laplace of 1/2(-sint-tcost) (we showed by convolution)
so y(t)= 1/2(-sint-tcost) + cost

 

[TFM]

Okay, So:

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

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

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

So:

Y = \frac{1}{2} (-sin(t)- tcos(t)) + cos(t)

Hows this?

TFM

 

[sara_87]

very good

but, don't say: s/(s^2+1) = cos(t)
say: s/(s^2+1)=laplace of cos(t)

 

[TFM]

So the final answer for this one is:

y(t) = \frac{1}{2} (-sin(t)- tcos(t)) + cos(t)

?

TFM

 

[sara_87]

yes.

 

[TFM]

Excellent. Sp the last one is very similar, except the values have changed slightly:

y&#039;&#039; + y = sin(t), y_0 = 1, y_0&#039; = -\frac{1}{2}

so:

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)


And again, there is no y',

L(y(t)) = Y

L(y&#039;(t)) = sL(y(t)) - 1

L(y&#039;&#039;(t)) = sL(y&#039;(t))-\frac{1}{2} = s^2(L(y(t))-sy(0)-y&#039;(0)

And again from before:

sin(t) = \frac{1}{p^2 + 1}

okay so far?

TFM