Differential Eq: Inverse Laplace Transforms

In summary, the conversation discusses two problems involving second-order differential equations and initial conditions. The first problem involves substitution and partial fractions, while the second problem requires completing the square in order to find the inverse using the properties of Laplace transforms. The final results for both problems are also mentioned.
  • #1
Gogeta007
23
0

Homework Statement



I have a couple of problems that I am stuck on. The following:

y'' - 6y' + 9y = t
y(0) = 0
y'(0) = 1

and

y'' - 6y' + 13y = 0
y(0) = 0
y'(0) = -3

Homework Equations



y'' = s2Y(s) - sf(0) - f'(0)
y' = sY(s) - f(0)
y = Y(s)


The Attempt at a Solution



===================================
for the first one:

once I substitute into the original equation, I can move things around and I came out with the following:
Y(s){s2-6s+9} = (1 +s2)/s2

So I get Y(s) = (1+s2)/s2(s-3)2

IIRC for partial fractions it should be the following: A/(s-3) + B/(s-3)2 + (Cs+D)/s2

I don't know if this is where i messed up but I got:
A=-2/27
B=-1/9
C=2/27
D=1/9

as a final result I get:
-2/27e^3t - 1/92e^3t and the other part (Cs + D)/s^2 . . .I can't find any way to transform that

the back of the book says:
1/9t + 2/27 - 2/27e^3t + 10/9te^3t

=============================

The second one:

Starting by substitution, plugging in the values and solving for Y(s)

Y(s) = -3/(s2-6s + 13)

and well. . I am lost from this point onwards. . .I don't remember how to do partial fractions if you can't factorize that denominator. And the quadratic formula (A=1 B = 6 C=13) gives imaginary numbers

=============================

Thank you!
 
Last edited:
Physics news on Phys.org
  • #2
Gogeta007 said:
as a final result I get:
-2/27e^3t - 1/92e^3t and the other part (Cs + D)/s^2 . . .I can't find any way to transform that.
You just have to separate it into individual terms.

[tex]\frac{Cs+D}{s^2} = \frac{C}{s} + \frac{D}{s^2}[/tex]

Also, according to Mathematica, you should have B=10/9. The other coefficients you found are correct.
 
Last edited:
  • #3
Gogeta007 said:
The second one:

Starting by substitution, plugging in the values and solving for Y(s)

Y(s) = -3/(s2-6s + 13)

and well. . I'm lost from this point onwards. . .I don't remember how to do partial fractions if you can't factorize that denominator. And the quadratic formula (A=1 B = 6 C=13) gives imaginary numbers.
Complete the square in the bottom so you get something that looks like

[tex]Y(s) = -\frac{3}{(s-a)^2+b^2}[/tex]

You should be able to find the inverse of that using the tables and the properties of Laplace transforms.
 
  • #4
wow. . .I can't believe I missed that. . .how does really old math comes back and haunt you huh?. . .thanks a lot
 

Related to Differential Eq: Inverse Laplace Transforms

What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that transforms a function from the complex frequency domain to the time domain. It is the inverse operation of the Laplace transform, which transforms a function from the time domain to the complex frequency domain.

What is the purpose of an inverse Laplace transform?

The purpose of an inverse Laplace transform is to solve differential equations in the time domain. By transforming a function from the complex frequency domain to the time domain, we can easily find the solution to a differential equation.

How do you perform an inverse Laplace transform?

To perform an inverse Laplace transform, you need to use a table of Laplace transform pairs or use a computer software. The process involves finding the corresponding function in the time domain by using the properties of the Laplace transform.

What are the properties of an inverse Laplace transform?

The properties of an inverse Laplace transform are linearity, derivative property, integral property, initial value theorem, final value theorem, and convolution property. These properties make it easier to solve differential equations by using the inverse Laplace transform.

What are some applications of inverse Laplace transforms?

Inverse Laplace transforms have various applications in engineering, physics, and mathematics. They are used to solve differential equations in control systems, signal processing, circuit analysis, and many other areas. They are also used in the study of heat transfer, fluid mechanics, and quantum mechanics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
858
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
208
  • Calculus and Beyond Homework Help
Replies
6
Views
900
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
266
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
923
  • Calculus and Beyond Homework Help
Replies
1
Views
690
  • Calculus and Beyond Homework Help
Replies
2
Views
476
Back
Top