Help with Laplace Transformations and 2nd order ODEs

[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