Laplace transformation initial value problem

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Homework Help Overview

The discussion revolves around solving an initial value problem (IVP) using Laplace transformations, specifically the equation y' - y = 2cos(5t) with the initial condition y(0) = 0.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transformation of the differential equation into the Laplace domain and the subsequent algebraic manipulation to find Y(s). There are attempts to clarify the setup of the equation and the correct form of partial fraction decomposition. Questions arise regarding the values of constants A and B, with some participants suggesting specific values for s to simplify calculations.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's work and clarifying misunderstandings. Some guidance has been offered regarding the correct setup of the equations and the identification of constants, but there is no explicit consensus on the final form of the solution yet.

Contextual Notes

Participants note the importance of correctly interpreting the initial problem statement and the implications of the initial conditions on the solution process. There is also a recognition of potential errors in earlier attempts, which are being addressed as the discussion progresses.

quietriot1006
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Homework Statement


I need to use the Laplace transformations to solve this IVP

y'-y=2cost(5t) , y(0)=0


Homework Equations





The Attempt at a Solution



I separated the question until the point

Y(s)=(2s)/(s^2+25)(s+1)

and then

2s=A(s+1)+B(s^2+25)

I get the answer for B=-(1/13) and i just want to know how to get A and if i am doing it right so far. ASAP would be great. Thanks.
 
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quietriot1006 said:

Homework Statement


I need to use the Laplace transformations to solve this IVP

y'-y=2cost(5t) , y(0)=0


Homework Equations





The Attempt at a Solution



I separated the question until the point

Y(s)=(2s)/(s^2+25)(s+1)

and then

2s=A(s+1)+B(s^2+25)

It would be helpful to see your work in how you got to the Y(s) equation. Also, I assume that the (s + 1) factor is in the denominator, but how you wrote it one could reasonable assume it is in the numerator.
For the 2nd equation above, did you separate the rational expression as A/(s^2 + 25) + B/(s + 1)? If so, that's incorrect. The first expression should be (As + B)/(s^2 + 25), and the other C/(s + 1)
quietriot1006 said:
I get the answer for B=-(1/13) and i just want to know how to get A and if i am doing it right so far. ASAP would be great. Thanks.
 
I messed up the first problem, there is no t after cos. This is how the problem is written.


y'-y=2cos(5t) , y(0)=0

Here is my work from the beginning:
sY(s)-0+Y(s)=(2s)/(s^2+25)

Y(s)=(2s)/((s^2+25)(s+1))

2s=((As+B)(s+1))+(C(s^2+25))

Let s=-1 and then i solve for C which comes out to equal (-1/13). And then I get stuck at A and B. I looked at what you said and you are right and i fixed that but I am still stuck.
 
Shouldn't you have sY(s)-Y(s)=2s/(s2+25) ?
 
Yes I should. Because of the minus from the initial problem. Thanks for that one. So
C=(1/13) with no minus sign. What about the rest? What number can I choose for s that will cause (s^2+25) to equal zero and then to get A and B?
 
try s=0 to get B
then put s=any number to get A
 
Cool! i think i got it.
y(t)=(-1/13)cos(5t)+(25/13)sin(5t)+(1/13)e^t

Is this right?
 

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