non-homogeneous constant co-efficient 2nd order linear differential equation

Question

I am doing a perturbation theory question and am having trouble with the (seemingly simple!) differential equation method of undetermined coefficients... I have reduced my given system so that now I need to find the solution of this non homogeneous differential equation: $$\ddot x+x=-cos(t) $$ with initial conditions:$$x(0)=1$$$$\dot x(0)=0$$ I'm so confused by this... I know that the solution is a combination of the general solution of the homogeneous equation plus a particular solution of the non homogeneous equation. I know the general solution is $$x_c=Acos(t)+Bsin(t)$$ but then I have to assume the particular solution for the homogenous equation is of the same form, $x_p=Ccos(t)+Dsin(t)$, since the right hand side of the original equation is in terms of cos. And when I find the second derivative of this and put it back into the equation, the terms cancel out until I am left with $0=-cos(t)$ which obviously doesn't help with finding the particular solution, and makes sense intuitively since I have just found out that this would be the general solution when it equals 0. Am I going about this all wrong or just missing something really obvious? Is 0 a particular solution to the non homogeneous equation? Any help would be greatly appreciated.

Answer 1

We are given:

$$\ddot x+x=-\cos(t) , ~x(0)=1, ~\dot x(0)=0 $$

For the homogeneous part of the system, we can write:

$$m^2 + 1 = 0 \implies m_{1,2} = \pm ~ i$$

The general solution is thus given by:

$$x_h(t) = c_1 \cos t + c_2 \sin t$$

Using the IC's, we find $c_1 = 1, c_2 = 0$, hence:

$$x_h(t) = \cos t $$

For the particular, since we have the solution equal to $\cos t$ and this is what we have for the complementary solution, we would choose:

$$\tag 1 x_p(t) = t(a \cos t + b \sin t)$$

If we did not have the particular solution coincident with the homogeneous, we would have chosen $x_p(t) = a \cos t + b \sin t$.

Taking derivatives of $(1)$ and subbing back into ODE yields:

$$-2 a \sin t + 2 b \cos t = - \cos t$$

This gives us:

$$a = 0, b = -\dfrac{1}{2}$$

The final solution is:

$$x(t) = x_h(t) + x_p(t) = \cos t - \dfrac{1}{2}~t~ \sin t$$