mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 134
Use the Laplace transform to solve the second-order initial value problem.
\(\displaystyle \L\\ y''-y = 2t\)
\(\displaystyle \L\\ y(0)= 0, y'(0)= -1\)
I use:
\(\displaystyle \L\\ L(y')(s)= sY(s)-y(0)\)
and
\(\displaystyle \L\\ L(y'')(s)= s^2Y(s)-sy(0)-y'(0)\)
I get
\(\displaystyle \L\\ Y(s)= {2}/{(s^2)(s^2\, -\, s)}\)
I factored it to be
\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)}\)
So, then I do partial fraction decomposition and get
\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)} = {A}/{s^3} + {B}/{s^2} + {C}/{s} + {D}/{(s-1)}\)
Now… is where I get lost. (That is assuming I did all of that right…)
\(\displaystyle \L\\ y''-y = 2t\)
\(\displaystyle \L\\ y(0)= 0, y'(0)= -1\)
I use:
\(\displaystyle \L\\ L(y')(s)= sY(s)-y(0)\)
and
\(\displaystyle \L\\ L(y'')(s)= s^2Y(s)-sy(0)-y'(0)\)
I get
\(\displaystyle \L\\ Y(s)= {2}/{(s^2)(s^2\, -\, s)}\)
I factored it to be
\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)}\)
So, then I do partial fraction decomposition and get
\(\displaystyle \L\\ {2}/{(s^3)(s\, -\, 1)} = {A}/{s^3} + {B}/{s^2} + {C}/{s} + {D}/{(s-1)}\)
Now… is where I get lost. (That is assuming I did all of that right…)