# Inverse Laplace Transfom - Two Solutions

Thread Starter

#### amilton542

Joined Nov 13, 2010
497
Because of the inverse Laplace transform thread that is already current in the math forum, I've decided to elaborate on this and show that there is a tendency for two different solutions to be generated.

Consider my example below. Forget about ODE's for now. For the first case I'll use partial fraction decomposition.

$$\frac{2s+3}{s^2 + 6s + 8} = \frac{2s + 3}{(s + 2)(s + 4)}$$

$$\frac{2s + 3}{(s + 2)(s + 4)} \equiv \frac{A}{(s+2)} + \frac{B}{(s+4)} \Rightarrow 2s + 3 \equiv A(s+4) + B(s+2)$$

$$A=-\frac{1}{2}\ and\ B=\frac{5}{2}$$

$$\frac{2s + 3}{(s + 2)(s + 4)} \equiv \frac{5}{2} \times \frac{1}{s+4} - \frac{1}{2} \times \frac{1}{s+2}$$

$$L^{-1}[ \frac{5}{2} \times \frac{1}{s+4} - \frac{1}{2} \times \frac{1}{s+2}] = -\frac{1}{2}e^{-4t}(e^{2t} - 5)$$

Intuition, however, says otherwise. This time I'll take the denominator and complete the square.

$$\frac{2s+3}{s^2 + 6s + 8} = \frac{2s+3}{(s+3)^2 - 1}$$

Now I'll add zero onto the end. There is nothing to stop me from doing this.

$$\frac{2s+3}{(s+3)^2 - 1} = \frac{2s+3}{(s+3)^2 - 1} + \frac{3}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1}$$

$$\frac{2s+3}{(s+3)^2 - 1} = \frac{(2)(s+3)}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1}$$

Taking the inverse Laplace transform,

$$L^{-1}[\frac{2s+3}{(s+3)^2 - 1}] = e^{-3t}\times[2ch(3t) - 3sh(t)]$$

Wolfram Alpha agrees with the former. But I cannot see what the problem is with the latter expression as a solution?

#### Papabravo

Joined Feb 24, 2006
16,072
The problems is that for the three cases of roots of the characteristic equation you have specific and unique formulations for the partial fraction expansion. The cases are:
Roots of the denominator are real and distinct.
Roots of the denominator are repeated.
Roots of the denominator are complex conjugate pairs.​
Using the complete the square method you have transformed case 1 (real and distinct) into case 2(real and repeated) and then dropped the ball.
The solutions to a linear differential equation are unique. Any different solution you get by whatever method is nonsense and is meaningless.

#### MrAl

Joined Jun 17, 2014
8,231
Hello,

As Papabravo says, if it doesnt work out then something is wrong.

How did you get that 'new' solution? Explain that and we can see what went wrong.

When i follow your work, no matter if i take the transform of the original, 'square' denominator, or the two new terms created with the 'zero' addition, i always get the same result:
5*e^(-4*t)/2-e^(-2*t)/2

Recall that the transform of the sum of terms is the sum of the transforms of the individual terms.

So perhaps if you show your work for getting your very last transform we can see what went wrong.

Thread Starter

#### amilton542

Joined Nov 13, 2010
497
@Papabravo

I don't understand how I've transformed the quadratic to have two repeated roots because there's still a constant tacked onto the end of it. Whenever I've wanted to curve sketch a quadratic I will always complete the square and I will always get a faithful sketch of the curve. Yes this can be risky. For example, if you don't watch your minuses there is a tendency to flip over the quadratic without even noticing.

If I set the quadratic to zero after completing the square I still get the two distinct roots. The quadratic formula IS derived from completing the square.

#### Papabravo

Joined Feb 24, 2006
16,072
You haven't really transformed it, it just looks that way. Something else is wrong if it leads to an alternate partial fraction expansion. I think you error is in taking the inverse Laplace transform. Can you show the steps that led to your result? The expansion and the inverse Laplace transform are unique and there is no getting around that. If you don't get the same expansion, the same inverse Laplace transform, and the same solution to the implied ODE then what you have created is wrong and has no real meaning or significance. In physical terms if there is more than one solution to an ODE then how does 'nature' pick which one it wants? The answer is that it doesn't have to pick because all solutions are unique.

Last edited:
Thread Starter

#### amilton542

Joined Nov 13, 2010
497
Apologies for the late response, I've been a bit pushed for time lately. I've spent a bit more time on my workings out for the second case and here they are. Please evaluate and criticize.

$$\frac{2s+3}{s^2 + 6s + 8} = \frac{2s+3}{(s+\frac{6}{2})^2 - (\frac{6}{2})^2 + 8} = \frac{2s+3}{(s+3)^2 - 1}$$

$$\frac{2s+3}{(s+3)^2 - 1} = \frac{2s+3}{(s+3)^2 - 1} + \frac{3}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1}$$

$$\frac{2s+3}{(s+3)^2 - 1} = \frac{2s+6}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1} = \frac{(2)(s+3)}{(s+3)^2 - 1} - \frac{3}{(s+3)^2 - 1}$$

$$L^{-1}\frac{2s+3}{(s+3)^2 - 1} = (2)L^{-1}\frac{(s+3)}{(s+3)^2 - 1} - (3)L^{-1}\frac{1}{(s+3)^2 - 1}$$

$$L^{-1}\frac{2s+3}{(s+3)^2 - 1} = 2e^{-3t}ch(t) - 3e^{-3t}sh(t) = e^{-3t}\times[2ch(t) - 3sh(t)]$$

#### Papabravo

Joined Feb 24, 2006
16,072
You need to review your basic algebra. You have not set up the undetermined coefficients.

Thread Starter

#### amilton542

Joined Nov 13, 2010
497
What do you mean? Please elaborate.

#### t_n_k

Joined Mar 6, 2009
5,455
Your solution in post #6 looks OK. It is equivalent to your initial solution.
One simply substitutes the Euler forms for cosh and sinh and the equivalence is obvious.

Thread Starter

#### amilton542

Joined Nov 13, 2010
497

#### MrAl

Joined Jun 17, 2014
8,231
Hello again,

This 'case' would have been 'solved' much sooner if you used standard notation for what you mean to state.
"sh" does not stand for the hyperbolic sine, and "ch" does not stand for the hyperbolic cosine.
The standard notation is:
cosh(x)
sinh(x)

If you had used this notation originally this would have been solved immediately because *everyone* here would have tried the hyperbolic equivalents and came up with the same transform as before. Instead it took 11 days, and you were lucky that member 'tnk' decided to try that or you still wouldnt know the answer Thread Starter

#### amilton542

Joined Nov 13, 2010
497
@MrAl

It's actually quite common to denote the hyperbolic sine and cosine using the shorthand notation. This should have been obvious to you when I took the inverse Laplace transform which took you 11 days to see that. Not my fault you couldn't get it right.

#### MrAl

Joined Jun 17, 2014
8,231
@MrAl

It's actually quite common to denote the hyperbolic sine and cosine using the shorthand notation. This should have been obvious to you when I took the inverse Laplace transform which took you 11 days to see that. Not my fault you couldn't get it right.
Hello,

Then how do you explain that it took 11 days for someone to investigate this by trying sinh and cosh ? And it wasnt me that investigated that after 11 days it was someone else wise guy.

Maybe more common in another language? Certainly not as common as sinh(x) and cosh(x) though for sure, which would have provoked a positive response the same day as you posted it ha ha.

Thread Starter

#### amilton542

Joined Nov 13, 2010
497
I would imagine t_n_k glanced the thread and pointed out it was a matter of a simple substitution by observing my final statement whereby I expressed the two hyperbolic functions in shorthand, hardly an investigation.

Here's a few more for you, just in case you come across these in the future.

cosec(x) = csc(x)

tanh(x) = th(x)

cosech(x) = csch(x)

#### Papabravo

Joined Feb 24, 2006
16,072
Well son of a gun. I told you they were the same thing even if it was difficult to recognize how they were the same. I guess if you set up a strawman you ought to be able to knock it down.

#### MrAl

Joined Jun 17, 2014
8,231
I would imagine t_n_k glanced the thread and pointed out it was a matter of a simple substitution by observing my final statement whereby I expressed the two hyperbolic functions in shorthand, hardly an investigation.

Here's a few more for you, just in case you come across these in the future.

cosec(x) = csc(x)

tanh(x) = th(x)

cosech(x) = csch(x)

Hello,

I'll agree that in the context sh() and ch() should probably have been taken to mean sinh() and cosh(), but also in the context you seemed to have a good grasp on what you were doing in the first place so i would assume that you would have tried that already. Since you stated that they were not the same, it is easy to reason that you went down that path already and it didnt work. So i could ask you why you didnt try that yourself to begin with.

In most texts we see:
sin(x), cos(x), tan(x), sinh(x), cosh(x), tanh(x), sec(x), csc(x).

In fact, of all the books i've read over the years they never used sh(x) and ch(x), and that includes engineering math handbooks and engineering handbooks, as well as countless data sheets.
But since you wrote the post and you knew for sure what they were, i have to wonder why you didnt try the subsitution yourself.

No problem though, as i see lots of abbreviations popping up on the internet these days. Some seem to be made up right on the spot and the writer expects all the readers to know what they mean So i'll concede and take your post as a good lesson in newer abbreviations, and in the mean time you can try to solve this equation:

y=t(x)+yu(em)-s(y)+ch(u)+c0-xfu(g)

[Just kidding on that one though ]

Good luck with your studies.

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