Linear differential equations: source term constant

[Morphheus]

Homework Statement

Solve the following differential equation for q(t) (position):

q''-qω^2 = C, where C is a time-independent value (basically a constant)

The Attempt at a Solution

This equation is not homogeneous, therefore it must be non-homogeneous.
However, in every definition of non-homogeneous differential equation I have found (textbooks and Internet), the source term (in this case, C) is labelled as dependent on time

So, do I apply the regular techniques to solve NH diff. equations? e.g. q(t) = qh(t) + qp(t) , where qh and qp are the homogeneous and particular solutions, respectively.

The only solution I may have would be to get rid of the constant by derivating both sides, and then solving q'''-q'w^2 = 0 instead, but I heavily doubt it's the right way.

 

[vela]

The constant function f(t)=C is still considered a function of time. You can apply the regular methods for finding the homogeneous and particular solutions.

You can also use your second approach of differentiating the entire equation and solving the resulting homogeneous equation, but you'll have to use the original equation to find the new arbitrary constant. This essentially amounts to solving the equation using the usual method.

 

[Morphheus]

You can also use your second approach of differentiating the entire equation and solving the resulting homogeneous equation, but you'll have to use the original equation to find the new arbitrary constant.

I don't really get the "using the original equation" part.

To solve such an equation, I would guess that q(t) = Ae^αt, then solve for alpha (which would give me α = ±√(something), and would finally re-plug these alphas using the superposition principle in q(t)

My final answer would be: Be^αt + Ce^-αt

 

[vela]

The equation q'''-ω²q'=0 has three roots, so you'll get three terms, each with an arbitrary constant. The solution to the original differential equation, however, should have only two arbitrary constants. To determine the third constant, you have to use the original differential equation.