- #1
Somefantastik
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In my notes, the prof wrote x' = x + sin(t), x(0) = x0
He did not go through the solution, only wrote that the solution is
x(t,x0) = x0et + et/2 - 1/2(sin(t) + cos(t))
I verified this is correct by substituting.
So when I try to solve this DE, here's what happens:
[tex] {\frac{dx}{dy} \ - \ x = \ sin(t) [/tex]
integrating factor is e-t
[tex] \int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt => [/tex]
[tex] e^{-t} \ x \ = \ \int e^{-t} \ sin(t) \ dt \ = \ -e^{-t} \ cos(t) \ - \ \int cos(t) \ e^{-t} \ dt \ = \ -e^{-t} \ cos(t) - \left( -e^{-t} \ + \int sin(t) \ e^{-t} \ dt \right) [/tex]
...
[tex] x = -(1/2)cos(t) + (1/2)sin(t) [/tex]
So I'm close, but where did I lose my e-t's ?
He did not go through the solution, only wrote that the solution is
x(t,x0) = x0et + et/2 - 1/2(sin(t) + cos(t))
I verified this is correct by substituting.
So when I try to solve this DE, here's what happens:
[tex] {\frac{dx}{dy} \ - \ x = \ sin(t) [/tex]
integrating factor is e-t
[tex] \int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt => [/tex]
[tex] e^{-t} \ x \ = \ \int e^{-t} \ sin(t) \ dt \ = \ -e^{-t} \ cos(t) \ - \ \int cos(t) \ e^{-t} \ dt \ = \ -e^{-t} \ cos(t) - \left( -e^{-t} \ + \int sin(t) \ e^{-t} \ dt \right) [/tex]
...
[tex] x = -(1/2)cos(t) + (1/2)sin(t) [/tex]
So I'm close, but where did I lose my e-t's ?