1st order non homogenous DE problem solving

[Somefantastik]

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:

{\frac{dx}{dy} \ - \ x = \ sin(t)

integrating factor is e-t

\int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt =>

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)

...

x = -(1/2)cos(t) + (1/2)sin(t)

So I'm close, but where did I lose my e-t's ?

[Ygggdrasil]

The solution to a non-homogeneous ODE is in the following form:

x = XH + Xsp

Where XH is the general solution to the corresponding homogeneous equation and Xsp 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 XH.

[HallsofIvy]

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:

{\frac{dx}{dy} \ - \ x = \ sin(t)

integrating factor is e-t

\int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt =>

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)

...

x = -(1/2)cos(t) + (1/2)sin(t)

So I'm close, but where did I lose my e-t's ?
You have e-x on both sides of the equation and they cancel out. Specifically, you get e-tx= -(1/2)e-t(cos(t)- sin(t)) by integration, then the "e-t" on each side cancel.

[Somefantastik]

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.

[Somefantastik]

You have found a specific solution, and you just need to solve the homogeneous equation dx/dt - x = 0 to obtain the XH.

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?

[Ygggdrasil]

Yes. Then you can apply your boundary conditions to get the final solution.

[Somefantastik]

Ok so x' = x would have to be x = et.

That gives me x = et - 1/2(sin(t) + cos(t)). It still does not match his final solution of

x(t,x0) = x0et + (1/2)et - 1/2(sin(t) + cos(t))

Any thoughts? I bet that I am misunderstand how to apply x(0) = x0

[Somefantastik]

For one thing I've been forgetting those constants from indefinite integration. I think my specific solution should look more like

x = -(1/2)cos(t) + (1/2)sin(t) + c(1/2) e^{t}


so I guess all that's left if helping me figure out how the x0et term got into the final solution:
x(t,x0) = x0et + (1/2)et - 1/2(sin(t) + cos(t))

[Ygggdrasil]

You forgot the constant in the solution to the homogeneous equation:
XH = Aet

Therefore the general solution to the non-homogeneous ODE is:
x(t) = Aet + (1/2)(sin(t)-cos(t))

To find the value for A, you should use the boundary condition x(0) = x0. (You should find that A = x0 + 1/2).


Alternatively, you could carry the constant through your original solution and solve for C using the boundary condition.

[Somefantastik]

Ok, I got it. Thanks for helping.