Differential Equation Using Laplace Transforms

[swordfish]

I need to solve the following differential equation using Laplace Transforms:

\(\displaystyle \frac{d^2s}{dt^2} + 9s = 2sin2t\)

Given that:

\(\displaystyle s=0\) and

\(\displaystyle \frac{ds}{dt} = 1\)

when \(\displaystyle t=0\)

I dont understand how to do the initial transform.

 

[Deleted member 4993]

I need to solve the following differential equation using Laplace Transforms:

\(\displaystyle \frac{d^2s}{dt^2} + 9s = 2sin2t\)

Given that:

\(\displaystyle s=0\) and

\(\displaystyle \frac{ds}{dt} = 1\)

when \(\displaystyle t=0\)

I dont understand how to do the initial transform.

This is a very "standard" problem of begining of Laplace transformation.

Your book should have example problems - also google and you'll find many example problems.

For an example problem - almost exactly like yours, go to:

http://www.sosmath.com/diffeq/laplace/a ... nswer.html


Please show us your work, indicating exactly where ypu are stuck - so that we know where to begin to help you.

 

[swordfish]

Am I right in assuming:

\(\displaystyle L(s) = 2 / (s^2+4)(s^2 +9)\)

which i should be able to slove using partial fractions?

 

[Deleted member 4993]

Am I right in assuming:

\(\displaystyle L(s) = 2 / (s^2+4)(s^2 +9) = 2\frac{s^2+9}{s^2+4}\) ...............incorrect

If you want me to confirm your answer - show work step-by-step

which i should be able to slove using partial fractions?

 

[swordfish]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+4}\)

\(\displaystyle s^2L(x) +9L(x) = \frac{2}{s^2+4} +1\)

\(\displaystyle L(x) = \frac{s^2 +6}{(s^2+4)(s^2+9)}\)

 

[Deleted member 4993]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+4} ..........................Incorrect\)

\(\displaystyle s^2L(x) +9L(x) = \frac{2}{s^2+4} +1\)

\(\displaystyle L(x) = \frac{s^2 +6}{(s^2+4)(s^2+9)}\)

\(\displaystyle L[sin(at)] \, = \, \frac{a}{s^2 + a^2}\)

 

[swordfish]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+2}\)

\(\displaystyle s^2L(x) +9L(x) = \frac{2}{s^2+2} +1\)

\(\displaystyle L(x) = \frac{s^2 +4}{(s^2+4)(s^2+9)}\)

 

[Deleted member 4993]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+2}\)
What are you doing???!!! ................?
\(\displaystyle s^2L(x) +9L(x) = \frac{2}{s^2+2} +1\)

\(\displaystyle L(x) = \frac{s^2 +4}{(s^2+4)(s^2+9)}\)
Does not follow from line above...

 

[swordfish]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+4}\)

\(\displaystyle s^2L(x) +9L(x) = \frac{2}{s^2+2} +1\)

\(\displaystyle s^2L(x) +9L(x) = \frac{2 + 1(s^2+2)}{s^2+2}\)

\(\displaystyle L(x) = \frac{s^2 +4}{(s^2+4)(s^2+9)}\)

Im far too confused now with what i am doing wrong

 

[Deleted member 4993]

\(\displaystyle \frac{d^2x}{dt^2} + 9x = 2sin2t\)

Given that: \(\displaystyle x=0\) and \(\displaystyle \frac{dx}{dt}=1\) when \(\displaystyle t=0\)



\(\displaystyle L(x'') + 9L(x) = 2L(sin2t)\)

\(\displaystyle L(x'') = s^2L(x) - sx(0) - x'(0) = s^2L(x) - 1\)

\(\displaystyle s^2L(x) - 1 +9L(x) = \frac{2}{s^2+4}\)
________________________________________
\(\displaystyle L[2sin(2t)] \, = \, \frac{2\cdot 2}{s^2 + 2^2} \, = \, \frac{4}{s^2 + 4}\)
________________________________________
\(\displaystyle s^2L(x) +9L(x) = \frac{4}{s^2+4} +1\)

\(\displaystyle s^2L(x) +9L(x) = \frac{4 + 1(s^2+4)}{s^2+4}\)

\(\displaystyle L(x) = \frac{s^2 +8}{(s^2+4)(s^2+9)}\)

Im far too confused now with what i am doing wrong

That is because you are not paying attention to what you are doing!!!