1st order non homogenous DE problem solving
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 ? 
Re: 1st order non homogenous DE problem solving
The solution to a nonhomogeneous 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 nonhomogeneous 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}. 
Re: 1st order non homogenous DE problem solving
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Re: 1st order non homogenous DE problem solving
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.

Re: 1st order non homogenous DE problem solving
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Re: 1st order non homogenous DE problem solving
Yes. Then you can apply your boundary conditions to get the final solution.

Re: 1st order non homogenous DE problem solving
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} 
Re: 1st order non homogenous DE problem solving
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)) 
Re: 1st order non homogenous DE problem solving
You forgot the constant in the solution to the homogeneous equation:
X_{H} = Ae^{t} Therefore the general solution to the nonhomogeneous 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. 
Re: 1st order non homogenous DE problem solving
Ok, I got it. Thanks for helping.

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