Laplace transformation to solve intial value problem

In summary: I think that I need to start with Y(s) and find the inverse laplace transform and then solve for U(s)...I am really lost at this point.In summary, the homework equations have the following:-y''(w-1)-2y'(w-1)+y(w+1)=4e-1-w-1+2(w+1)U(w)=y(w+1) when w=0and u'(w)=y'(1)=-3 when w=0
  • #1
Karmel
12
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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...
 
  • #6
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
 
  • #7
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...
 

1. What is a Laplace transformation?

A Laplace transformation is a mathematical tool used to convert a function of time into a function of complex frequency. It is commonly used in engineering and physics to solve differential equations and initial value problems.

2. How is a Laplace transformation used to solve initial value problems?

A Laplace transformation is used to convert a differential equation into an algebraic equation in the complex frequency domain. This algebraic equation can then be solved using various techniques, such as partial fraction decomposition, to obtain the solution in the time domain.

3. What types of initial value problems can be solved using Laplace transformation?

Laplace transformation can be used to solve linear initial value problems, which involve a linear differential equation and initial conditions. It may not be suitable for solving non-linear problems or problems with discontinuous functions.

4. What are the advantages of using Laplace transformation to solve initial value problems?

Laplace transformation can simplify the process of solving initial value problems, as it converts a differential equation into an algebraic equation. It also allows for the use of algebraic techniques, which may be easier to work with than differential equations.

5. Are there any limitations to using Laplace transformation to solve initial value problems?

Although Laplace transformation is a powerful tool, it may not be suitable for solving all types of initial value problems. It may also be more complex and time-consuming than other methods, such as the method of undetermined coefficients or variation of parameters.

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