Laplace transform of sawtooth function for 2nd order ode

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spcrooks
spcrooks am 25 Nov. 2018
Kommentiert: Aquatris am 26 Nov. 2018
Hi all,
I have taken the rhs laplace for a sawtooth equation where:
f(t)=2t for 0<t<1
and f(t+1)=f(t)
T=2
>> syms s t lapf
lapf =simplify(int('exp(-s*t)*2*t','t=0 .. 2')/(1-exp(-s)))
pretty(lapf)
lapf =
-(2/s^2 - (2*(2*s + 1))/(s^2*exp(2*s)))/(1/exp(s) - 1)
Now I need to solve the differential equation: y'' + y = f(t)
Here, f(t) is the sawtooth function above.
I am having some difficulty marrying the two. Any suggestions would be appreciated.
Thank you!
  5 Kommentare
spcrooks
spcrooks am 26 Nov. 2018
I thought about your initial post earlier, and that led me to this...
>> syms s t Y
f = f(t);
F = laplace(f,t,s);
Y1 = Y;
Y2 = Y*s^2;
sol = solve(Y2 + Y1 - F, Y);
However, now I need to figure out how to appropriately insert f(t), which sawtooth function from the OP. I have tried the code with a dummy value for "f" and it works.
Aquatris
Aquatris am 26 Nov. 2018
If thats what you want to do, and you are sure about the laplace transform of your sawtooth function, then the answer is easy;
Y = -(2/s^2 - (2*(2*s + 1))/(s^2*exp(2*s)))/(1/exp(s) - 1)/(s^2+1)
You already determined F in your question (the variable lapf), so why are you confused?

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