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From: sean on 5 Mar 2010 04:34
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






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