Question

Solve this initial value problem...
\y''-y=t-2 \space ; \space y(2)=3 \space , \space y'(2)=0\]

Answer 1

\[y''-y=t-2 \space ; \space y(2)=3 \space , \space y'(2)=0\]

Answer 2

Time shift...
w(t)=y(t+2)
w'(t)=y'(t+2)
w''(t)=y''(t+2)

Answer 3

Check me so far please

Answer 4

@zepdrix

Answer 5

\[w''-w=t \space ; \space w(0)=3 \space , \space w'(0)=0\]

Answer 6

After partial fractions I end up with...\[\large -\frac{ 1 }{ s^2 }+3\frac{ s }{ s^2-1 }-\frac{ 1 }{ s^2-1 }\]

Answer 7

inverse Laplacing I get\[-t+3\cos(-t)-\sin(-t)\]

Answer 8

Back of the book says\[2-t+e^{2-t}+2e^{t-2}\]

Answer 9

I am missing the final step looks like. I have to reverse the time shift

Answer 10

Hmm I'm not familiar with the time shift thing.. (Just starting to learn this stuff past friday). Is it because we need our initial conditions to start at time 0?

Answer 11

Ya. Initial conditions must be at zero for Laplace

Answer 12

if w(t)=y(t+2), then y(t)=w(t-2). It looks like I might have made a mistake during the problem. I shouldn't have any trig in here. I'm going to rework it. I'll post the work if you're interested to see a time shift @zepdrix

Answer 13

Mmm ok i think i see your mistake, sec lemme check my work a sec.