Help with Laplace Transformations and 2nd order ODEs

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

 

[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 separately 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.

 

[TFM]

Okay so:

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

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

So, the Laplace tranformation of:

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

and the Laplace transformation of:

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

And finally, the Laplace Transformation of:

$$\frac{1}{2s^2 + 2}$$

would this be similar to the first one, but require first shift theorem:

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

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

Is this okay?

TFM

 

[sara_87]

The first two bits are perfectly fine.
what is the inverse laplace transform of 1/(s^2 +1)
It is ofcourse sin(t)
so that means the inverse laplace of 1/2(s^2+1) is just simply (1/2)sin(t)

so now you have the inverse laplace of each bit, now you can state what the function y(t) is.

 

[TFM]

Okay, so now:

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

Is this okay?

TFM

 

[sara_87]

that's right

 

[TFM]

Excellent. Thanks for all your assistance, sara_87