Question

Use Laplace transforms to solve the following initial value problem:
(dy/dt)− y = e^t ... y(0) = 1

Answer 1

guess what, you need to laplace transform both sides of the equation first.

Answer 2

yeah i get that tommy...but its the end part that gets me. i get ... (s/(s-1)^2) but now i gotta invert it to find for y but i cant simplify my answer to do so.

Answer 3

Hold on, I haven't done this in while :)

Answer 4

okay, thanks :)

Answer 5

you need to make two fractions of s/(s-1)^2 like this: \[\frac{s}{(s-1)^{2}}=\frac{A}{s-1}+\frac{B}{s-1}\]
So you need to figure out A and B.

Answer 6

A(s-1)+b(s-1)=s

Answer 7

Actually it doesn't work like that for this particular fraction. wolfram alpha says: \[\frac{s}{(s-1)^{2}}=\frac{1}{s-1}+\frac{1}{(s-1)^2}\]

Answer 8

Now you can invert it.

Answer 9

s/(s-1)^2 Where does this come from, please explain...)

Answer 10

A(s-1)+b(s-1)=s

Answer 11

After the laplace transform you get:\[sY-1-Y=\frac{1}{s-1}\]
with a little algebra you get \[Y=\frac{s}{(s-1)^2}\]

Answer 12

I didn't do the algebra part ;)

Answer 13

Are u supposed to just read them from a list..?

I had 1-1/s = 1/s+1 (probably I am reading it wrong)

Answer 14

You can read them from a table, or memorize them, or use the definition to find transform. But these are pretty standard.

Answer 15

Assume I know nothing....

Answer 16

I have a table here, is it the right one..?

Answer 17

http://en.wikipedia.org/wiki/Laplace_transform

Answer 18

So u(t) = y = 1/s, is that right?

Answer 19

well that's what wikipedia says, I'm confused about the u(t) part.

Answer 20

I was assuming that u is just a dummy for x or y or whatever...

It's how you go from the equation to

sY-1-Y= 1/s-1 I don't get..

Answer 21

the laplace transfrom of all those things in the table is the same without u(t).

Answer 22

anyway, Y is the laplace transform y.

Answer 23

It's unknown at this point.

the laplace transfrom of the derivate of a function is s*(the laplace transfrom of that function)-f(0)

Answer 24

and the laplace transform of e^t is 1/(s-1)

Answer 25

So in the table it says 1/s+alpha....

Answer 26

where alpha is coefficient in e^at so in this case 1, no?

Answer 27

-1

Answer 28

So the table is wrong?

Answer 29

Can u direct me to a good table?

Answer 30

the table is right, it says e^-at=1/s+a, so you need -1 for a to e^t.

Answer 31

Ah, blind......:-)

Answer 32

ok, that's the right hand side, how about he left?

Answer 33

Well you can take the laplace transform each of the terms seperately.

Answer 34

so L(y)=Y, L(dy/dt)=sY-y(0)=sY-1

Answer 35

Aah...so you end up with Y in terms of s, I see now...

Answer 36

right, then you can transform back to t.

Answer 37

Thank u muchly:-)

Answer 38

you're welcome :)

Answer 39

Thanks to both of you. I tried the same partial fraction as you Thomas9. Your other method cleared it up nicely. Thanks a million!!!