Diff EQ with the boundary conditions

[Warr]

I took an ODE course last year, but I seem to have forgotten some stuff. I need to solve this equation:

\frac{d^2u}{dt^2} + {\omega}^2u = f_osin({\mu}t)

with the boundry conditions:

u(0) = 0, du/dt(0) = 0

When I tried to solve the homogenenous equation first, I got

u_g(t)=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}

I then differentiated and set up the system with the two boundy conditions...but I got c1+c2=0 and c1-c2=0...c1=c2=0.

This seems wrong. Any help would be appreciated

 

[AKG]

I think those might be called initial conditions, not boundary conditions, not that it matters. Anyways, you need to find a particular solution, add it to your u_g, then find the c_i by looking at the initial conditions.

 

[Warr]

Here is what I did now:

let the particular solution be of the form u_p=Asin({\mu}t})

differentiated twices gives u_p''=-A{\mu}^2sin({\mu}t})

so now from my origional diff eq I get

A({\omega}^2-{\mu}^2)sin({\mu}t)=f_osin({\mu}t)

henceA = \frac{f_o}{{\omega}^2-{\mu}^2}

and therefore

u_p=\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})

so now the general solution is

u=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})

plugging in the initial conditions from the first post, I got

c_1=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}

c_2=-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}

and therefore the entire solution to be

u=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{i{\omega}t}-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})

 

[AKG]

I think you're missing a factor of i\omega in the denominators of c₁ and c₂.

 

[Warr]

You are right. Also, is it right to say that the solution exists as long as |\mu| {\neq} |\omega| (assuming that they are both real)?

Thanks for the help!

 

[HallsofIvy]

If you are going to use sin(\mu t) as you particular solution, why not use sin(\omega t) and cos(\omega t) in the general solution?

Yes, the solution is what you have as long as |\mu|\ne|\omega|. However, if you are thinking that when they are equal, a solution does not exist, that is not necessarily so. You just need to use a more complicated function as your specific solution.