Laplace transform of sawtooth function for 2nd order ode

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spcrooks
spcrooks 2018-11-25
评论： Aquatris 2018-11-26
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!
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spcrooks
spcrooks 2018-11-26
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 2018-11-26
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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