Laplace Transforms (First Order DE with Initial Value)

In summary, the problem involves solving for y in the equation Y' + 8y = e^-2t*sint, with an initial condition of y(0) = 0. The solution involves using partial fraction decomposition to solve for the coefficients A, B, and C, and the final result is y(t) = (-1/37)e^-2t * (sin(t) + 8cos(t)) + (4/37)e^-8t * sin(t). The discrepancy mentioned in the conversation may be due to a typing error in the book.
  • #1
cas159
5
0
Y' + 8y = e^-2t*sint, with initial condition y(0) = 0

Alright so I've been working on this one for about an hour, I really don't know why but I'm having major problems with these types of problems, whether i don't understand how to set it up or I don't understand partial fraction decomposition, I do not know, but this is what I have done.

(For reference, L{function} is the notation I will use).

L{y''} + 8L{y} = L{e^(-2t)*sin(t)}

sY(s) - y(0) + 8Y(s) = 1/((s+2)² + 1)

1/((s+2)² + 1) = 1/(s²+4s+5)

(s+8)Y(s) = 1/(s²+4s+5)

Y(s) = 1/((s²+4s+5)(s+8))

Now setting up my Partial Fraction Decomposition

As+B/(s²+4s+5) + C/(s+8) = 1/((s²+4s+5)(s+8))

As² + Bs + 8As + 8B + Cs² + 4Cs + 5c = 1

(A+C)s² + (8A + B + 4C)s + 8B + 5C = 1

Setting up my equations:

A + 0B + C = 0
8A + B + 4C = 0
0A + 8B + 5C = 1

I'm getting that A = -1/37, B = 4/37, and C = 1/37, the rest I can do, but the book says that B should be 6/37, can anyone figure out what I did improperly, I would really appreciate it.
 
Physics news on Phys.org
  • #2
I agree with B=4/37
May be a typing mistake in the book. Did you compare the final results y(t) ?
 
  • #3
I did, and with a friend as well and he got 4/37 as well. Thanks a lot, books are so awesome sometimes.
 
  • #5
Okay sorry, I posted it in the one and read the rules and posted in the correct one (the homework one), am I allowed to delete a post? But yea, sorry.
 

Related to Laplace Transforms (First Order DE with Initial Value)

What is a Laplace transform?

A Laplace transform is a mathematical operation that converts a function of time into a function of frequency. It is commonly used in solving differential equations, particularly first-order differential equations with initial values.

How is a Laplace transform used to solve first-order differential equations with initial values?

A Laplace transform can be used to convert a first-order differential equation into an algebraic equation, making it easier to solve. The initial values can then be used to find the specific solution to the differential equation.

What are the advantages of using Laplace transforms in solving differential equations?

One of the main advantages of using Laplace transforms is that they can simplify complex differential equations into easier to solve algebraic equations. They also allow for the use of initial values to find specific solutions, which may not be possible with other methods.

Are there any limitations to using Laplace transforms in solving differential equations?

While Laplace transforms can be a powerful tool in solving differential equations, they may not always be applicable. They are most useful for linear differential equations with constant coefficients, and may not work for non-linear equations or equations with variable coefficients.

Can Laplace transforms be used in real-world applications?

Yes, Laplace transforms have many real-world applications in fields such as engineering, physics, and economics. They are commonly used in modeling and analyzing systems that involve time-varying processes, such as electrical circuits and mechanical systems.

Similar threads

  • Differential Equations
Replies
17
Views
884
  • Differential Equations
Replies
1
Views
705
  • Calculus and Beyond Homework Help
Replies
7
Views
833
  • Calculus and Beyond Homework Help
Replies
1
Views
120
  • Differential Equations
Replies
11
Views
1K
  • Differential Equations
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Differential Equations
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
164
Back
Top