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DE using Laplace transform

  • Thread starter manenbu
  • Start date
  • #1
103
0

Homework Statement



y'' -6y' + 9y = f(t)

f(t)=
0, 0<t<1
1, 1<t<3
0, t>3

Homework Equations





The Attempt at a Solution



Turning it to heaviside functions I get:
y'' -6y' + 9y = u1 - u3

and I solve.

in the answers it should be:
y(t) = 2tet + u1(1/9 - 1/9e3(t-1)+ 1/3(t-1)e3(t-1)).

where did u3 go in the answers? why doesn't it appear? the u1 I can get just like that, but I can't seem to figure out where the u3 went and why there is a 2tet in there.
 

Answers and Replies

  • #2
41
0
It looks like we're missing some information. Were there additional conditions given? (e.g. y(0) = ? and y'(0) = ?)

Also, what did the DE look like after you applied the Laplace transform?
 
  • #3
1,033
1
it doesn't look as though you changed anything by putting it in unit step form.
 
  • #4
103
0
Sorry, forgot.
y(0) = 1, y'(0) = 2.
 

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