Solving a differential equation in distribution theory Let $$a_n\frac{d^ny(t)}{dt^n} + a_{n-1}\frac{d^{n-1}y(t)}{dt^{n-1}} + \dots +a_0y(t) = \delta(t)$$What's the general solution for $y(t)$? I don't know how to interpret and solve this equation in the sense of distribution theory. Certainly $\delta(t)$ is not a function, so it can't be evaluated at point. I'm looking for a rigorous solution which gives general solution. Here is a specific example which is solved by md2perpe but doesn't seem can be generalized to the general case.
 A: If you take the solution $y_0$ with initial values $0=y_0(0)=y_0'(0)=...=y_0^{n-2}(0)=0$ and $y_0^{n-1}(0)=1$, then you can confirm that $y(x)=y_0(x)u(x)$, with the unit ramp function $u$, inserted into the differential operator will give a right side of
$$
(n-1)!a_n\delta(t).
$$
Now divide $y$ by the unwanted factors.
A: Especially if we want to behave as though there were "methods" beyond "tricks", we might want to see how to solve the differential equation in a way that genuinely uses distributions. (I'm not going to say anything here that's not available in some form in the references linked-to by Wikipedia's page on the Malgrange-Ehrenpreis theorem, though the "more elementary arguments" that were published in the Math Monthly do not seem to be widely known.)
Let the equation be $P(i\partial)u=\delta$, where $\partial=d/dx$, and $P$ is a complex polynomial. When $P$ has no real roots, using Fourier transform and simply "dividing" suffices. The opposite case is more interesting: suppose $P(x)=0$ has $n$ distinct real roots $a_1,\ldots,a_n$. We still use Fourier transform, but cannot naively "divide".
Supposing $u$ is at least a tempered distribution, take the Fourier transform: $P(x)\widehat{u}=1$. We would like to divide, but the functions $1/(x-a_j)$ are not locally $L^1$, so are not immediately (tempered) distributions. But we might recall that $xv=1$ for distribution $u$ implies that $v$ is a constant multiple of the principal_value integral against $1/x$. At least qualitatively and heuristically, $1/P(x)=\sum_j c_j/(x-a_j)$ for suitable constants $c_j$, so it is reasonable/plausible that the tempered distribution $v$ such that $P(x)\cdot v=1$ is the corresponding linear combination of principal value integrals. (And this is true, but does take a little proof, which will vary depending on one's set-up.)
We also know that the Fourier transform of PV $1/x$ is (up to a constant) the sign function. Fourier transform converts translation to multiplication by exponentials. Thus, the (inverse) Fourier transform is (a constant multiple of) $\sum_j c_j\cdot ({\mathrm sign}(x))\cdot e^{ia_jx}$.
To check that this is correct requires a little more info about the coefficients $c_j$, for example, that $c_j=1/P'(a_j)$. This uses the distinctness of the $a_j$.
