Why Does My Laplace Transform Calculation Differ from the Textbook's Result?

In summary, the student attempted to solve a homework equation, but got stuck. They were assisted by someone who was able to help them find the answer using integration by parts.
  • #1
Eastonc2
20
0

Homework Statement


f(t)=e^(t+7)

Homework Equations


£{f(t)}=∫e^(-st)f(t)dt

The Attempt at a Solution


so i insert my f(t) into the formula, came up with ∫e^(-st+t+7)dt
using u substitution, u=t(-s+1)+7, du=(-s+1)dt so it follows that 1/(-s+1)∫e^(u)du=e^(u)/(-s+1)
so I plug u back in, and should be able to find my answer from there, only I come up with an answer very different from the one in the book, which is e^(7)/(s-1)
Can anyone help me out?

So i figured it out, I set u=-1(t(-s+1)+7)=t(s-1)-7, and put e^-u inside the integral. turns out just making myself look at it a little longer worked out
 
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  • #2
Eastonc2 said:

Homework Statement


f(t)=e^(t+7)


Homework Equations


£{f(t)}=∫e^(-st)f(t)dt


The Attempt at a Solution


so i insert my f(t) into the formula, came up with ∫e^(-st+t+7)dt
using u substitution, u=t(-s+1)+7, du=(-s+1)dt so it follows that 1/(-s+1)∫e^(u)du=e^(u)/(-s+1)
so I plug u back in, and should be able to find my answer from there, only I come up with an answer very different from the one in the book, which is e^(7)/(s-1)
Can anyone help me out?

You didn't show your work with the limits, which is where your error is. You need to either put the new u limits in your u answer or the t limits in your t answer.
 
  • #3
ok, so I figured out that last one, now I'm having difficulty with £{t^(2)e^(-2t)}. putting it into the laplace definition I come up with ∫t^(2)e^(-t(s+2))dt. I've tried integration by parts, and come up with:
u=t^2, du=2t dt, dv=e^(-t(s+2))dt

and that's where i get stuck, i can't seem to figure out this integration. I've plugged it into wolfram, but that turns out with v=te^(-t(s+2)), which, when plugged back into the integration by parts, leaves me with a more complicated equation, involving the negative of my original integral.
 
  • #4
You can always use integration by parts again,
 
  • #5
changed a few things around, still using integration by parts, actually integrated by parts i think three times total, if I'm remembering last night correctly, to finally end up with the right answer. thanks
 

FAQ: Why Does My Laplace Transform Calculation Differ from the Textbook's Result?

What is the Laplace transform and what is it used for?

The Laplace transform is a mathematical tool used to transform 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 with complex inputs.

What are the advantages and disadvantages of using the Laplace transform?

One advantage of using the Laplace transform is that it can simplify complex differential equations and make them easier to solve. It also allows for the use of algebraic manipulations instead of calculus. However, one disadvantage is that it may not always have a unique inverse, making it difficult to obtain the original function.

How is the Laplace transform calculated?

The Laplace transform is calculated by taking the integral of a function multiplied by the exponential of a negative variable. This results in a new function in the frequency domain. The inverse Laplace transform is then used to convert the function back to the time domain.

What is the difference between the one-sided and two-sided Laplace transforms?

The one-sided Laplace transform is used for functions that are only defined for positive time values, while the two-sided Laplace transform is used for functions that are defined for both positive and negative time values.

What are some common applications of the Laplace transform?

The Laplace transform has a wide range of applications, including electrical networks, control systems, signal processing, and fluid mechanics. It is also used in probability and statistics to analyze random processes and in economics to solve differential equations related to economic models.

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