Can't seem to get this DE solved using laplace transforms

In summary, laplace transforms are helpful in solving differential equations because they convert them into simpler algebraic equations. However, common mistakes when using laplace transforms include forgetting initial conditions, not accounting for inverse laplace transform, and discontinuities. Laplace transforms are most effective for linear equations with constant coefficients and may have limitations for non-linear or higher order equations. To check the accuracy of the solution, one can take the inverse laplace transform and compare it to the original equation or plug it into the original equation to see if it satisfies initial conditions.
  • #1
1MileCrash
1,342
41

Homework Statement



y''-4y'+4y = 0
y(0)=1, y'(0)=1

Homework Equations





The Attempt at a Solution



This is annoying, because it feels very easy and I don't know why I am not getting the answer. Skipping the routine:

s^2L{y} - s - 1 - 4sL{y} + 4 + 4L{y} = 0

(s+3) / (s^2 - 4s + 4) = L{y}

= s/(s-2)^2 + 3/(s-2)^2

= 1/(s-2) + 2(1/(s-2)^2) +3(1/(s-2)^2)

= 5(1/(s-2)^2) + 1/(s-2)


The inverse transform of which gives:

5te^2t + e^2t = (5t+1)e^2t


Any ideas?

Thanks!
 
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  • #2
NVM, I think it should be s-3 in num now.
 
  • #3
Check your signs.

EDIT: Ah, I see you found the mistake.
 

1. Why is using laplace transforms helpful in solving differential equations?

Laplace transforms can help simplify and solve complex differential equations by converting them into algebraic equations, which are easier to manipulate and solve.

2. What are some common mistakes when using laplace transforms to solve differential equations?

Some common mistakes include not properly setting up the initial conditions, forgetting to take into account the inverse laplace transform, and not accounting for any discontinuities in the original function.

3. Can laplace transforms be used to solve any type of differential equation?

Laplace transforms are most effective for solving linear differential equations with constant coefficients. They may not be as useful for non-linear or higher order differential equations.

4. Are there any limitations to using laplace transforms in solving differential equations?

One limitation is that laplace transforms can only be used for equations with continuous functions. They also may not work for equations with varying coefficients or with certain types of boundary conditions.

5. How can I check if my solution using laplace transforms is correct?

You can check your solution by taking the inverse laplace transform and comparing it to the original differential equation. You can also plug your solution into the original equation and see if it satisfies the initial conditions.

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