1. The problem statement, all variables and given/known data Solve the following problems using Laplace Transforms: [tex] y' - y = 2e^t, y_0 = 3 [/tex] [tex] y'' + 4y' + 4y = e^{-2t}, y_0 = 0, y_0' = 4 [/tex] [tex] y'' + y = sin(t), y_0 = 1, y_0' = 0 [/tex] [tex] y'' + y = sin(t), y_0 = 1, y_0' = -\frac{1}{2} [/tex] 2. Relevant equations N/A 3. The attempt at a solution I don't seem to understand how to use Laplace Transformation to solve ODE's! If anyone can help, I will be most grateful. TFM
lets look at the first one: let L(F(t)) denote the laplace of F(t) in general, you say that: L(y'(t))-L(y(t))=L(2e^t) call this (1) L(y(t))=y (in terms of s) L(y'(t))=sL(y(t) - y(0) (This is one of the formulas from the table of laplace transforms) =sy-3 so, substituting these back into (1): sy-3-y=2/(s-1) (also from the tables) make y the subject, then you should get an expression for y in terms of s. Then to find y(t) (which is what you are looking for), you must find the inverse of y. I hope all is clear.
It might be confusing to use the same label y for the function und its Laplace transform. Better come up with a new letter like Y You then have to solve (according to Sara) [tex]sY(s)-3-Y(s)=\frac{2}{s-1}[/tex] for Y(s) and find the inverse Laplace transform of this function.
Okay lets see if this makes sense: you have the orginal equation, F(t) [tex] y' - y = 2e^t [/tex] then you make L(f(t)): [tex] L(y'(t)) - L(y(t)) = L(2e^t) [/tex] ---(1) I still can't see how you have gone from here to here: ??? TFM
that is not from one step to another. L(y'(t))-L(y(t))=L(2e^t) this is something and L(y(t))=Y is something else. In other words, LET L(y(t))=Y so therefore, L(y'(t))=sY-3
Let's see then...: [tex] L(F(t): L(y'(t)) - L(y(t)) = L(2e^t) [/tex] ---(1) and also: [tex] L(y(t)) = Y [/tex] so inserting into (1) [tex] L(F(t): L(y'(t)) - Y = L(2e^t) [/tex] [tex] L(y'(t)) = Y + L(2e^t) [/tex] I see now where the 3 comes from, [tex] y_0 = 3 [/tex] [tex] L(y'(t)) = Y - y_0 [/tex] [tex] L(y'(t)) = Y - 3 [/tex] substitute back into (1): [tex] Y - 3 - L(y(t)) = L(2e^t) [/tex] does this look right so far? TFM
you made a small mistake, L(y'(t))=sY-3 because you must use the formula: L(F'(t))=sL(F(t))-F(0) Im not sure but I think you are expected to know this general formula. so this gives: sY-3-L(y(t))=L(2e^t)
Yes, the formula: [tex] L(F(t)) = sL(F(t)) - F(0) [/tex] looks useful. I assume the [tex] sL(F(t)) [/tex] is inequivalent to: [tex] s*L(y(t)) = Y [/tex] ??? so: [tex] L(y'(t)) = sY - 3 [/tex] putting back into (1) [tex] L(y'(t)) - L(y(t)) = L(2e^t) [/tex] [tex] sY - 3 - sY - 3 = L(2e^t) [/tex] does this look right? TFM
The star means convolution (in Laplace language) so it is inequivalent You have: L(y'(t))-L(y(t))=L(2e^t) this is (1) we said let L(y(t))=Y and then we know that L(y'(t))=sY-3 so putting into (1) gives: sY-3-Y=L(2e^t) Why are you putting sY-3 twice?
I meant the star as in times, Sorry. I am not sure, how am I putting in sY-3 twice? sY-3-Y=L(2e^t) only has sY - 3 in once? TFM
exactly so again you have to remember: L(y'(t))-L(y(t))=L(2e^t) this is (1) (A) we said let L(y(t))=Y (B) and then we know that L(y'(t))=sY-3 so putting into (1) gives: (C) sY-3-Y=L(2e^t) These are the three main steps that should give you the equation of Y in terms of s. So, what would you do next?
Well, judging from your first post, rearrange into Y? so this would be: [tex] sY-3-Y=L(2e^t) [/tex] put 3 to one side: [tex] sY-Y=L(2e^t) + 3 [/tex] factorise out Y: [tex] Y(s - 1) = L(2e^t) + 3 [/tex] divide by (s - 1): [tex] Y = \frac{L(2e^t) + 3}{s - 1} [/tex] Does this look right? ??? TFM
no, before you 'put to one side' on your second step, you have to find the laplace of 2e^t sY-3-Y=L(2e^t) so, sY-3-Y=2/(s-1) (do you know that L(2e^t)=2/(s-1) ?) now, you make Y the subject.
I have a table with Laplace conversions, and it gives: [tex] e^{-\alpha t} = \frac{1}{p + \alpha} [/tex] so [tex] 2e^{t} = \frac{2}{p + t} [/tex] I'm assuming p and s are equivilant so now: [tex] sY-3-Y = L(2e^t) [/tex] [tex] sY-3-Y = \frac{2}{s + t} [/tex] now take over the 3: [tex] sY-Y = \frac{2}{s + t} + 3 [/tex] factorise Y: [tex] Y(s - 1) = \frac{2}{s + t} + 3 [/tex] thus giving Y [tex] Y = \frac{\frac{2}{s + t} + 3}{s - 1} [/tex] Does this look better? TFM
e^(-at)=1/(s-a) so: (the right hand side is the laplace so it has nothing to do with t) 2e^(t)=1/(s+1) (since in our case, -a=1) Now, make Y the subject from: sY-3-Y=1/(s+1) (do you agree?)
I feel slightly sill, I meant to do that, but didn't So: [tex] sY-3-Y = \frac{2}{s + 1} [/tex] now: [tex] sY-Y = \frac{2}{s + 1} + 3 [/tex] [tex] Y(s - 1) = \frac{2}{s + 1} + 3 [/tex] [tex] Y = \frac{\frac{2}{s + 1} + 3}{s - 1} [/tex] Is this right? TFM
Im sorry, i misread; the L(2e^t) means that our a=1 so that means L(2e^t)=2/(s-1) not 2/(s+1) but your steps are correct and so your Y would be: sY-3-Y=2/(s-1) Y(s-1)=2/(s-1) +3 so Y=2/(s-1)^2 + 3/(s-1) (by dividing both sides by s-1) is this clear? so, what would you do next?
so: [tex] Y = \frac{\frac{2}{s - 1} + 3}{s - 1} [/tex] is the same as: [tex] Y = \frac{2}{(s - 1)^2}+\frac{3}{s - 1} [/tex] so now, from your first post again, I need to find the inverse: thus: [tex] \frac{1}{Y} = \frac{(s - 1)^2}{2}+\frac{s - 1}{3} [/tex] ??? TFM
no, carefull.... when i say inverse i dont mean 1/Y; Y=2/(s-1)^2 + 3/(s-1) call this (2) (This is you Laplace of the function you are looking for) So you want to find the function in terms of t that would give you (2) when you find its laplace. so you want to find a function of t. so, Y(t)= something that when you find its laplace you would get (2) so, first, what is the function in terms of t that when you find its laplace you would get 2/(s-1)^2 ? (you would need to use your tables)