Laplace transformation to solve intial value problem

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Homework Help Overview

The discussion revolves around applying the Laplace transformation to solve an initial value problem involving a second-order differential equation. The equation presented is y'' - 2y' + y = -4e^(1-t) + 2t, with initial conditions y(1) = 0 and y'(1) = -3.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the Laplace transformation to both sides of the equation and the subsequent steps to isolate Y(s). There are attempts to clarify the transformations of derivatives and the handling of initial conditions. Some participants express uncertainty about the direction to take after finding the Laplace transforms.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and questioning specific steps in the transformation process. Some guidance has been offered regarding the transformations and the need to adjust variables to align with initial conditions. However, there is no explicit consensus on the next steps or the correctness of the approaches taken.

Contextual Notes

Participants note the need to change variables to ensure initial conditions are applied correctly, which introduces additional complexity to the problem. There is also mention of specific Laplace transform properties that are being discussed but not fully understood by all participants.

Karmel
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Homework Statement


laplace transformation to solve intial value problem
y"-2y+y=-4e^1-t + 2t
y(1)=0, y'(1)=-3


Homework Equations





The Attempt at a Solution


K. I have one more on the review that I am unaware of where I need to start. So if anyone can point me in the right direction please feel free. My graduation depends on the final and I am so lost on how to start amny of these problems... Thanks in advance
 
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Take the Laplace transformation on both sides noting that L{y(t)}=Y(s), then just make Y(s) the subject, then find the inverse laplace transform of Y(s) and you will the solution.
 
okay so I found the laplace of both sides and get

2/s^3 + 2/s^2 + 1/s = (-4e(-t+1)/s) + 2t/s

I'm not really sure where to go form here. I think that I want to set
Y=L{y(t)}(s)
L{y'}=sY
L{y"}=s^2Y

but I am not so sure that is the direction I need to head and I really am not sure were all this fits into the problem. Where do I head from here?
 
L{y(t)}=sY(s)-Y'(0)

L{y''(t)}=s2Y(s)-sY(0)-Y'(0)

L{ekt}= 1/(s-k)

L{tn}=n!/sn+1
 
ok I get the first two but why the second two in the above reply. I guess I don't know what I need to do with them...
 
On looking back to the question, you need to change the variable such that the conditions occur when t=0. To do this we let t=w+1

so now we have

y''(w-1)-2y'(w-1)+y(w+1)=4e-1-w-1+2(w+1)

Now define u(w)=y(w+1)
so now u'(w)=y'(w+1) and u''(w)=y''(w+1)

Now when w=0 => u(0)=y(0+1)=y(1)=0 AND u'(w)=y'(1)=-3

so now you need so solve the new equation

u''-2u'+u=4e-w+2w+2


Now take the Laplace transform on both sides of the equation.

to elaborate on the second two...


if you need to get L{e5t} it is simplt 1/(s-5)

and L{t2}=2!/s3
 
Okay so now I get
L{u"}(s) = s2U(s) + 3
L{u'}(s) = sU(s)


u''-2u'+u=4e-w+2w+2

taking the LT of both sides I get
s2U(s)+3-2sU(s)+U(s)=(4/s+1)+(2/s2)+(2/s)
(s2-2s+1)U(s)=(4/s+1)+(2/s2)+(2/s)-3

I feel like I have messed up somewhere cause if I solve for U(s) then it gets really ugly...
 

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