LaplaceTransform for integrodifferential equation with unit step
- To: mathgroup at smc.vnet.net
- Subject: [mg108018] LaplaceTransform for integrodifferential equation with unit step
- From: sean <sean_incali at yahoo.com>
- Date: Fri, 5 Mar 2010 04:33:56 -0500 (EST)
Hello Group,
I can solve the following integro-differential equation using
LaplaceTransform.(This one doesn't have unit step)
In[17]:=
LaplaceTransform[10 i[t] + 0.01 i'[t] + 1600\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)]\(i[z] \
[DifferentialD]z\)\) ==5, t, s]
%//Expand
%/. LaplaceTransform[i[t],t,s]-> i[s]
Solve[%, i[s]]/. i[0]-> 0//Flatten//ExpandAll//Factor
Out[20]=
{i[s]->500./((200.+s) (800.+s))}
Using partial fractions, we find that the laplace transform i[s].
In[31]:=
Solve[(200.`+s) (800.`+s)==0,s]
i[s]==500.`/((200.`+s) (800.`+s))==a/(s+200)+b/(s+800)
500.==a (800.`+s)+b (200.`+s)//Expand
{a s+b s==0,800 a+200 b==500}//Solve[#,{a,b}]&
i[s]==a/(200+s)+b/(800+s)/.%
InverseLaplaceTransform[%, s, t]/.InverseLaplaceTransform[i[s],s,t]-
>i[t]
Out[36]=
{i[t]==5/6 E^(-800 t) (-1+E^(600 t))}
Now if I add the step by defining
u[t] := Piecewise[{{0, t < 0}, {1, 0 < t <= 1}}, {t, -\[Infinity], \
[Infinity]}]
it's basically a 1 second pulse u[t]
And try to solve the resulting equation, it doesn't seem to work.
LaplaceTransform[10 i[t] + 0.01 i'[t] + 1600 \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(t\)]\(i[
z] \[DifferentialD]z\)\) == 5 u[t], t, s]
% // Expand
% /. LaplaceTransform[i[t], t, s] -> i[s]
Thank you all for any insights in advance.
Sean