Using laplace transforms to solve IVPplease check work thanks

In summary, the given differential equation can be solved by using the Laplace transform method and decomposing the denominator to find the inverse Laplace transform. Some algebra mistakes were made in the solution attempt, resulting in an incorrect answer, but upon correcting these mistakes, the solution matched the answer provided in the book. Additionally, it was noted that even if the denominator is not fully decomposed, the resulting solution should still be equivalent to the solution found when the denominator is fully decomposed.
  • #1
fufufu
17
0

Homework Statement


y' - 3y = 13cos(2t)

y(0)=1

Homework Equations


y' = sY(s) - y(0)

The Attempt at a Solution



heres all my work.. i am confused as to why its not matching book solution.. i think (geussing) that I probably messing up the decomposition step..thanks for any help with this
https://docs.google.com/open?id=0BwJqUg33PgREVjdEN250WXVTaldod3NublhVc1pEdw

Homework Statement


Homework Equations


The Attempt at a Solution


sorry almost forgot to include this..book solution is:

y(t) = 4e^3t - 3cos(2t) + 2sin(2t)
 
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  • #2
You made a couple of algebra mistakes, one of them due to an omission of parentheses where they were needed.
 
  • #3
thanks for the help..yup i see it: (As+B)(s-3) makes all the difference, giving me C=3, B=4 and A = -3...now matches book solution.. but have follow-up question..

If I start solving equation with what I think is a fully decomposed denominator, when denominator could actually be decomposed further, then the answer I get should be identical to the one solved with fully decomposed denomiator, right? so, even though i probably had to solve with more steps because deenominator wasnt fully decomposed, the answers shouls still be the same, true?
 
  • #4
Yes, you should ultimately get an equivalent answer. The two results may not be expressed in exactly the same way, but they will be equal to each other.
 

Related to Using laplace transforms to solve IVPplease check work thanks

What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function from the time domain to the frequency domain. It is commonly used in engineering and physics to solve differential equations and analyze systems.

How can Laplace transforms be used to solve IVPs?

Laplace transforms can be used to solve initial value problems (IVPs) by transforming the differential equation into an algebraic equation in the frequency domain. The solution can then be easily found by taking the inverse Laplace transform of the transformed equation.

What are the advantages of using Laplace transforms to solve IVPs?

One advantage of using Laplace transforms is that they can be used to solve linear differential equations with constant coefficients, which may be difficult to solve using traditional methods. Additionally, Laplace transforms can handle discontinuous or piecewise functions, making them useful for real-world applications.

Are there any limitations to using Laplace transforms to solve IVPs?

One limitation of using Laplace transforms is that they can only be used to solve linear differential equations. Nonlinear equations, such as those involving products or powers of the dependent variable, cannot be solved using Laplace transforms. Additionally, the initial conditions for the IVP must be known in order to use this method.

Can Laplace transforms be used to solve IVPs with variable coefficients?

Yes, Laplace transforms can be used to solve IVPs with variable coefficients. However, the method becomes more complex as the coefficients vary with time. In these cases, partial fraction decomposition and inverse Laplace transforms may be required to find the solution.

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