Help doing differential equations and laplace

In summary: So if you want to invert the Laplace transform of a function, you just need to find the poles at the negative real part of the function's argument and use the residue theorem.
  • #1
jbhungal
1
0
I have to solve and IVP we were given an 2x2 matrix and i used the determinant method to solve for L=s^2 +2s+2, Lx =s-2 and Ly =s. so now i did this LX=Lx so s-2/(s^2+2s+2) and LY=Ly so s/(s^2+2s+2) now i have to Laplace them and I am stuck. I think i have to make the (s^2+2s+2) => ((s+1)^2 +1) but I am not sure bc it doesn't really help
 
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  • #2
I have no idea what you mean by "so s-2/(s^2+ 2x+ 2)" or "so s/(s^2+ 2s+ 2)". Are those equal to something? I also have no idea what you mean by "Laplace them". Do you mean "find their inverse Laplace transform"? Are you trying to find the inverse Laplace transform of each separately or are you combining them somehow?
 
  • #3
If you want to find the inverse Laplace transform of
[tex]\frac{s-a}{(s+1)^2+1} [/tex]then find the poles which are at [tex]s=-1 \pm i [/tex] so the sum of the residues of
[tex] e^{sx}\frac{(s-a)}{(s+1-i)(s+1+i)}[/tex] are: [tex] e^{sx}\frac{(s-a)(s+1-i)}{(s+1-i)(s+1+i)}|_{s=-1+i}+ e^{sx}\frac{(s-a)(s+1+i)}{(s+1-i)(s+1+i)}|_{s=-1-i}=

e^{(-1+i)x}\frac{(-1+i-a)}{2i}-e^{(-1-i)x}\frac{(-1-i-a)}{2i}
[/tex]

[tex]
=
2*Real (e^{-x}(cos(x)+isin(x))\frac{i(1+a)+1}{2})

=e^{-x} (cos(x)-(1+a)sin(x))
[/tex]

So the inverse Laplace transform of [tex]\frac{s}{s^2+2s+2}[/tex] is [tex]e^{-y} (cos(y)-sin(y))[/tex] and of [tex]\frac{s-2}{s^2+2s+2}[/tex] is [tex]e^{-x} (cos(x)-3sin(x))[/tex].

The reason this is so is that the contour integral for x>0 vanishes at a large semicircle at the real part of s at -infinity and so the inversion formula [tex]f(x)=\frac{1}{2\pi i} \int F(s)e^{sx}ds [/tex] picks out [tex]2\pi i*residues[/tex].
 

Related to Help doing differential equations and laplace

1. What is the purpose of using differential equations and Laplace transforms in scientific research?

Differential equations and Laplace transforms are fundamental tools used in scientific research to model and understand the behavior of complex systems. They help to describe how variables change over time, and can be used to predict future outcomes and make informed decisions.

2. What are some common applications of differential equations and Laplace transforms?

Differential equations and Laplace transforms have a wide range of applications in various fields such as engineering, physics, biology, and economics. Some common examples include analyzing electrical circuits, studying population growth, and predicting the motion of celestial bodies.

3. How do you solve differential equations using Laplace transforms?

To solve a differential equation using Laplace transforms, the equation is first transformed into the Laplace domain. This involves applying the Laplace transform to both sides of the equation. The resulting algebraic equation can then be solved using various techniques, such as partial fraction decomposition or the inverse Laplace transform.

4. What are some challenges in using differential equations and Laplace transforms?

One of the main challenges in using differential equations and Laplace transforms is the complexity of the equations involved. Many real-world problems require the use of multiple differential equations and complex Laplace transforms, which can be difficult to solve analytically. Additionally, choosing the appropriate initial conditions and boundary conditions can also be challenging.

5. Are there any alternatives to using differential equations and Laplace transforms?

Yes, there are alternative methods for solving problems that involve differential equations. Some examples include numerical methods such as Euler's method and Runge-Kutta methods, which use a series of approximations to find a solution. Additionally, there are also other types of transforms, such as Fourier transforms, that can be used for similar purposes.

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