In my notes, the prof wrote x' = x + sin(t), x(0) = x_{0} He did not go through the solution, only wrote that the solution is x(t,x_{0}) = x_{0}e^{t} + e^{t}/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 ?
The solution to a non-homogeneous ODE is in the following form: x = X_{H} + X_{sp} Where X_{H} is the general solution to the corresponding homogeneous equation and X_{sp} is a specific solution to the given non-homogeneous ODE. You have found a specific solution, and you just need to solve the homogeneous equation dx/dt - x = 0 to obtain the X_{H}.
You have e^{-x} on both sides of the equation and they cancel out. Specifically, you get e^{-t}x= -(1/2)e^{-t}(cos(t)- sin(t)) by integration, then the "e^{-t}" on each side cancel.
I understand they cancel; my question referred to the fact that the prof's answer included e to the POSITIVE t and my answer included no e^t's at all.
so I would solve dx/dt - x = 0 the same way that I solved dx/dt = x + sin(t), and then I should add that result with the particular solution and that is my general solution?
Ok so x' = x would have to be x = e^{t}. That gives me x = e^{t} - 1/2(sin(t) + cos(t)). It still does not match his final solution of x(t,x_{0}) = x_{0}e^{t} + (1/2)e^{t} - 1/2(sin(t) + cos(t)) Any thoughts? I bet that I am misunderstand how to apply x(0) = x_{0}
For one thing I've been forgetting those constants from indefinite integration. I think my specific solution should look more like [tex] x = -(1/2)cos(t) + (1/2)sin(t) + c(1/2) e^{t} [/tex] so I guess all that's left if helping me figure out how the x_{0}e^{t} term got into the final solution: x(t,x_{0}) = x_{0}e^{t} + (1/2)et - 1/2(sin(t) + cos(t))
You forgot the constant in the solution to the homogeneous equation: X_{H} = Ae^{t} Therefore the general solution to the non-homogeneous ODE is: x(t) = Ae^{t} + (1/2)(sin(t)-cos(t)) To find the value for A, you should use the boundary condition x(0) = x_{0}. (You should find that A = x_{0} + 1/2). Alternatively, you could carry the constant through your original solution and solve for C using the boundary condition.