Solving a 2nd order ODE

[emmingja]

I've been trying to wrap my head around the step taken just under Eq 22 in these instructions: http://www.maplesoft.com/content/EngineeringFundamentals/49/MapleDocument_46/Orbita_Mechanics.pdf

I have an ODE of the form x'' + x = C. I'm trying to solve for x.

They say:
"Solve the ODE by adding the particular solution, x_p = C, to the general solution x=Acos(theta + theta_o) with theta_o = 0."
so they get: x = Acos(theta) + C

I do not understand this. I thought it should be x = Acos(theta) + Bsin(theta) + C.
Other sources say that the Bsin(theta) = 0 because of initail conditions. C

Can someone please explain to me what the initial conditions are that result in x = Acos(theta) + C

Thank you!!

Jason

 

[Ishuda]

I've been trying to wrap my head around the step taken just under Eq 22 in these instructions: http://www.maplesoft.com/content/EngineeringFundamentals/49/MapleDocument_46/Orbita_Mechanics.pdf

I have an ODE of the form x'' + x = C. I'm trying to solve for x.

They say:
"Solve the ODE by adding the particular solution, x_p = C, to the general solution x=Acos(theta + theta_o) with theta_o = 0."
so they get: x = Acos(theta) + C

I do not understand this. I thought it should be x = Acos(theta) + Bsin(theta) + C.
Other sources say that the Bsin(theta) = 0 because of initail conditions. C

Can someone please explain to me what the initial conditions are that result in x = Acos(theta) + C

Thank you!!

Jason

The general solution to the homogeneous equation is
x = A sin(\(\displaystyle \theta\)) + B sin(\(\displaystyle \theta\))
Letting
C = \(\displaystyle \sqrt{A^2 + B^2}\),
\(\displaystyle sin(\theta_0) = \frac{A}{C}\)
\(\displaystyle cos(\theta_0) = \frac{B}{C}\)
we can write this as
x = A sin(\(\displaystyle \theta)\) + B cos(\(\displaystyle \theta\)) = C [\(\displaystyle sin(\theta) sin(\theta_0) + cos(\theta) cos(\theta_0)\)]
Or
x = C cos\(\displaystyle (\theta - \theta_0)\).
Just after Eq. (22) is the initial condition \(\displaystyle \theta_0\) = 0 and the solution to the homogeneous solution is
x = C cos\(\displaystyle (\theta)\).
Adding the particular solution, x is constant, to the general homogeneous solution with the initial condition of \(\displaystyle \theta_0\) = 0 and changing what we are calling the constants, that is A is changed to C, and the x is constant is the constant C, we finally have
x = A cos\(\displaystyle (\theta)\) + C.

ta daaaaaaaaaaaa!

 

[Ishuda]

Oh, and note that the initial condition of \(\displaystyle \theta_0\) = 0, is the same as the original A = 0.