Laplace transform of an expression using transform tables

PainterGuy · Apr 20, 2020

Hi,

I 'm trying to find the Laplace transform of the following expression.

I used the following conversion formulas.
I think "1" is equivalent to unit step function who Laplace transform is 1/s.

Two Laplace transform table entries: e^(-at) sin(ωt) transforms to ω/((s+a)^2+ω^2), and e^(-at) cos(ωt) transforms to (s+a)/((s+a)^2+ω^2).

I ended up with the following final Laplace transform.

Handwritten-style algebra in red showing a chain of equalities simplifying 1/s - (1/3)*3/((s+1)^2+3^2) - (s+1)/((s+1)^2+3^2) down to (s^2+s+10)/(s(s^2+2s+10)).

Is my final result correct? Thank you for the help!

DrClaude · Apr 20, 2020

Your Laplace transforms are correct, but not your subsequent algebra. The last term in the first (red) line is where the error appears.

PainterGuy · Apr 20, 2020

DrClaude said:

The last term in the first (red) line is where the error appears.

I tried to fix the error. Could you please give it a look? For some reason, I think that in the numerator I should only get "10" and not "2s+10".

$Multi-line algebra working with fractions in s, simplifying 1/s minus (1/3)·3/((s+1)^2+3^2) minus (s+1)/((s+1)^2+3^2) down to (2s+10)/(s(s^2+2s+10)), with intermediate steps shown in red.$

DrClaude · Apr 20, 2020

You make the same mistake in the numerator, now at the beginning of the second line.

PainterGuy · Apr 20, 2020

DrClaude said:

You make the same mistake in the numerator, now at the beginning of the second line.

Sorry!

It looks good now.

Red typeset algebra in several lines simplifying a Laplace transform: 1/s minus (1+s+1)/((s+1)^2+3^2), then to (s+2)/(s^2+2s+10), combined over s(s^2+2s+10) to reach 10/(s(s^2+2s+10)).

DrClaude · Apr 20, 2020

Correct!

Laplace transform of an expression using transform tables

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