Help with Laplace Transformations and 2nd order ODEs

[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:

$$y''+ y = sin(t)$$

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:

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

Can be simplified to:

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

Which can go to:

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

Better?

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:

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

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',

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

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

$$L(y''(t)) = sL(y'(t))-\frac{1}{2} = s^2(L(y(t))-sy(0)-y'(0)$$

And again from before:

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

okay so far?

TFM

 

[sara_87]

yes that's right. now substitute everything in and make Y the subject

 

[TFM]

So original equation:

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

L(y(t)) = Y

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

$$L(y''(t)) = s^2(Y)-1-1/2 = s^2(Y) - 3/2$$

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

Still okay?

TFM

 

[sara_87]

first, the -1/2 should be 1/2 since you have: --1/2=1/2, 2nd:
you made the same mistake again:
in the second to last step, you have -1 but it should be -s
L(y(t))=Y
L(y'(t))=sY-1
L(y''(t))=s(sY-1)+1/2=s^2Y-s+1/2

now substitue and make Y the subject

 

[TFM]

first, the -1/2 should be 1/2 since you have: --1/2=1/2, 2nd:
you made the same mistake again:
in the second to last step, you have -1 but it should be -s
L(y(t))=Y
L(y'(t))=sY-1
L(y''(t))=s(sY-1)+1/2=s^2Y-s+1/2

now substitue and make Y the subject

I think I've found why I'm making the same mistake. Should this:

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

Really be:

$$L(y'(t)) = s(L(y(t)) - 1)$$

It would work then?

Anyway, so:

L(y(t))=Y

L(y''(t))=s(sY-1)+1/2=s^2Y-s+1/2

$$y'' + y = sin(t)$$

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

rearrange for Y:

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

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

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

Is this okay?

TFM

 

[sara_87]

very good.
now, let's separate them to make life much simpler:
Y=1/(s^2+1)^2 + s/(s^2+1) - 1/2(s^2+1)
now find the inverse laplace of each one seperately and remember we found the inverse laplace of the first fraction before using convolution so you don't have to do it again but u could for practice if you wish.