Inverse Laplace Transform, can't use partial fractions

In summary, the conversation discusses a problem involving a second order differential equation and its Laplace transform. The equation is solved using partial fractions, and the inverse Laplace transform is calculated using complex roots. The final solution is (3/14)*exp(-t/3)*sin(wt), where w = sqrt(14)/3.
  • #1
_diego
2
0
Hi, this is my first post, and I'm not sure if I'm posting this in the right place.

Homework Statement



I have a 2nd order diff. equation:
3y''(t) + 2y'(t) + 5y(t) = 3
with initial values: y'(0) = 1, y(0) = 0


Homework Equations


After using Laplace transform I get:

Y(s) = (3 + 3s) / (s*[3s2 +2s +5])

I believe it is correct, but even if it's not, what I'm interested in is how to solve this particular inverse laplace transform where I can't use partial fractions due to 3s2 +2s +5 not having any roots.

The Attempt at a Solution


I'm kind of stuck here, but maybe I could split the fraction like:

Y(s) = 3 / (s*[3s2 +2s +5]) + 3 / ([3s2 +2s +5])

but the problem would remain.

Thanks
 
Physics news on Phys.org
  • #2
You could split it as

A/s + (Bs + C)/(3s2 + 2s + 5)
 
  • #3
I would us an online solver such as WolframAlpha type in:

DSolve[{3y''[t]+2y'[t]+5y[t]==3,y'[0]==1,y[0]==0},y[t],t] and enter
 
  • #4
Re-reading my post I think I didn't explain my problem correctly. My problem is in calculating the inverse transform of such rational function.
I could split it like that, but then I would have no idea on how to solve:
L-1[ (Bs + C)/(3s2 + 2s + 5) ]

Since I'm studying for a final exam, I really can't use a solver, but thanks.
 
  • #5
Thanks for the clarification. Looking at the WolframAlpha solution can still be helpful. Their solution looks very similar to a homogeneous second order diff. eqn with complex roots. Might be a place to start.
 
  • #6
_diego said:
Hi, this is my first post, and I'm not sure if I'm posting this in the right place.

Homework Statement



I have a 2nd order diff. equation:
3y''(t) + 2y'(t) + 5y(t) = 3
with initial values: y'(0) = 1, y(0) = 0


Homework Equations


After using Laplace transform I get:

Y(s) = (3 + 3s) / (s*[3s2 +2s +5])

I believe it is correct, but even if it's not, what I'm interested in is how to solve this particular inverse laplace transform where I can't use partial fractions due to 3s2 +2s +5 not having any roots.

The Attempt at a Solution


I'm kind of stuck here, but maybe I could split the fraction like:

Y(s) = 3 / (s*[3s2 +2s +5]) + 3 / ([3s2 +2s +5])

but the problem would remain.

Thanks

Of course you can use partial fractions. The function 3s^2 + 2s + 5 has the form 3*(s-r1)*(s-r2), where r1 and r2 are the two roots of the quadratic; they are complex: r1 = -(1/3) + (1/3)*sqrt(14)*I and r2 = -(1/3) - (1/3)*sqrt(14)*I, where I = sqrt(-1). Then 1/[(s-r1)(s-r2)] = [1/(s-r1) - 1/(s-r2)]/(r1 - r2). The inverse Laplace transform of 1/(3s^2+2s+5) is [(1/3)/(r1-r2)]*[exp(r1*t) - exp(r2*t)] = (1/3)*exp(-t/3)*[(2i)sin(w*t)]/(2i*sqrt(14)/3) = (3/14)*exp(-t/3)*sin(wt), where w = sqrt(14)/3.

R.G. Vickson
 

Related to Inverse Laplace Transform, can't use partial fractions

What is the purpose of an Inverse Laplace Transform?

The Inverse Laplace Transform is used to find the original function from its Laplace Transform. It is the reverse process of the Laplace Transform, which is used to convert a function from the time domain to the frequency domain.

Why can't partial fractions be used in the Inverse Laplace Transform?

Partial fractions can only be used when the Laplace Transform of a function is in the form of a ratio of polynomials. However, not all Laplace Transforms can be expressed in this form, which is why partial fractions cannot always be used in the Inverse Laplace Transform.

How can the Inverse Laplace Transform be calculated without using partial fractions?

The Inverse Laplace Transform can be calculated using other methods such as the residue theorem, convolution integral, or the use of tables and properties of Laplace Transforms. These methods are often used when partial fractions cannot be used.

What are some common mistakes to avoid when using the Inverse Laplace Transform?

Some common mistakes to avoid when using the Inverse Laplace Transform include incorrect application of the inverse formula, not taking into account initial conditions, and not using the correct properties of Laplace Transforms. It is important to carefully follow the steps and double-check calculations to avoid these mistakes.

Can the Inverse Laplace Transform be used for all functions?

No, the Inverse Laplace Transform can only be used for functions that have a Laplace Transform. Some functions, such as non-analytic functions, do not have a Laplace Transform and therefore cannot have an Inverse Laplace Transform.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
411
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
195
  • Electrical Engineering
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Back
Top