Why are differential equations with sinusoidal source terms easier to solve than others? I am a software engineer trying to wrap my tiny human brain around Fourier Transforms for a project I'm currently working on. Although I will ultimately use an open source Math library to do all the heavy lifting for me, I don't like to do anything without at least having a basic understanding of it, so I came here.
My understanding is that linear differential equations with non-sinusoidal source terms are hard to solve (though I don't understand why). And so Fourier Transform helps convert these problems into several component linear differential equations with sinusoidal source terms, which are apparently easy to solve.
So first, if I have misunderstood the motivation/reasoning behind the use of Fourier Transforms, please begin by correcting me!

Assuming I'm more or less correct, there are a few mental blockers for me here:


*

*Does Fourier Transform only apply to linear differential equations? If so, why?

*How many smaller "component" functions (with sinusoidal source terms) does Fourier Transform produce?


Moreover, why (from a 30,000 ft view) is solving something like this (don't get hung up on the specific functional definitions, I'm just providing these as straw men examples):
L[y(t)] = 3t^2 + 4t + 8          // Non-sinusoidal source term

harder to solve than something like this:
L[y(t)] = 2sin(4*PI + t) - 20    // Sinusoidal source term

? Thanks for any-and-all clarification/help.
 A: That's because exponentials are "eigenfunctions" of linear differential equations with constant coefficients. They spontaneously appear in homogenous equations and fit naturally in non-homogeneous ones.
The basic reason is the property
$$(e^{at})'=ae^{at}$$
telling that the derivative of an exponential is the same exponential, times a constant. When you plug that in a differential equation, say
$$y''(t)+Ay'(t)+By(t)=0,$$ you get the ordinary equation
$$(a^2+Aa+B)e^{at}=0$$
which is verified by the roots of the (so-called characteristic) polynomial.
In stable systems, the root(s) $a$ are negative real numbers or complex numbers with a negative real part. The function
$$e^{(-1/\tau+i\omega)t}=e^{-t/\tau}(\cos(\omega t)+i\sin(\omega t)),$$
is a sinusoid with exponential decay.
Now if you consider a sinusoidal RHS, let $Ce^{bt}$, you known that by plugging that function in the LHS you will get a similar exponential:
$$(b^2+Ab+B)e^{bt}=Ce^{bt}.$$
The solution is simply
$$\frac C{b^2+Ab+B}e^{bt}.$$

This property of the (complex) exponentials is so advantageous that a common resolution method is to decompose the RHS as a (possibly infinite) sum of exponentials, solve the equation for every component and recombine the partial solutions. This is the working principle of the Fourier/Laplace transform approach.
