Distribution theory - Find a solution to the linear partial differential equation $\partial u+au=\phi$ This was a problem on my final last semester. I am trying to learn how this is done for the future. I thought that I figured it out and I emailed the instructor and he said there was a simpler solution without using Fourier transforms. I was wondering if anyone knows what to do as I am very stuck...

What I did:
We want a solution of the form $$LE=\delta$$ where $\delta$ is the delta function.
Then, $$u=E * \phi$$
So, we have 
$$\partial E +aE = \delta$$ This is where I took the Fourier transform to get,
$$-i\xi\hat{E}+a\hat{E}=1$$
$$\Rightarrow (a-i\xi)\hat{E}=1$$
$$\Rightarrow \frac1{(a-i\xi)}+c\delta=\hat{E}$$ is a family of solutions. Then we can take $c=0$. 
Then I have $$E=\int_{-\infty}^\infty \frac{e^{-i\xi x}}{a-i \xi} \frac{d\xi}{2\pi}$$
Again, I was told there was a much simpler solution. I am hoping someone can help me out.
Thanks.
 A: Using the method of integrating factor:
$$E'+aE = \delta$$
$$(Ee^{ax})' = \delta e^{ax} = \delta$$
$$Ee^{ax} = H$$
$$E = He^{-ax},$$
where $H$ is the Heavyside function.
A: The equation $\partial u + au = \phi$ is functionally equivalent to the ODE $u'(x)+au(x)=\phi(x)$. Since we are only required to find a solution, we will take the additional condition $u(x_0) = 0$ (this can be relaxed but it makes the calculations messier). We look for a fundamental solution $E(x;\xi)$ which satisfies $E'(x;\xi)+aE(x;\xi) = \delta(x-\xi)$ and $E(x_0;\xi) = 0$ for all $\xi>x_0$. When $x\neq\xi$, $\delta(x-\xi) = 0$, so $E(x;\xi) = \begin{cases} c_<\exp(-ax) & x_0\leq x<\xi \\ c_> \exp(-ax) & \xi<x \end{cases}$
Integrating the ODE on the interval $(\xi-\epsilon,\xi+\epsilon)$ gives us
$$ \int_{\xi-\epsilon}^{\xi+\epsilon} E'(x;\xi)\text dx + a\int_{\xi-\epsilon}^{\xi+\epsilon}E(x;\xi)\text dx = \int_{\xi-\epsilon}^{\xi+\epsilon}\delta(x-\xi)\text dx $$
The second term vanishes and the RHS $\to 1$ as $\epsilon\to 0$. The first term is $E(\xi+\epsilon;\xi)-E(\xi-\epsilon;\xi)$, which gives us the size of the jump discontinuity of $E$ at $x=\xi$, we'll call it $J_E(\xi)$. Then, the above integral expression becomes
$$ J_E(\xi) + 0 = 1 $$
So, this jump condition gives us the constraint $c_> \exp(-a\xi) - c_< \exp(-a\xi) = 1$. This, combined with the initial condition, gives us two constraints with which we can solve for the two constants.
By the initial condition, $c_< \exp(-ax_0) = 0$, so $c_< = 0$. This reduces the second equation to $c_> \exp(-a\xi) = 1$, so $c_> = \exp(a\xi)$. Plugging these constant back in, we get an expression for the fundamental solution
$$ E(x;\xi) = \begin{cases} 0 & x_0\leq x<\xi \\ \exp(-a(x-\xi)) & \xi<x \end{cases} = H(x-\xi)\exp(-a(x-\xi)) $$
where $H$ is the Heaviside function. We notice that $E(x;\xi) = F(x-\xi)$ with $F(t) = H(t)\exp(-at)$.
Thus, the solution to the original equation can be written as
$$ u(x) = \int_{x_0}^\infty \phi(\xi)E(x;\xi)\text d\xi = (\phi * F)(x)$$
(Note: your definition of $*$ may require $x_0 = 0$, and as such that particular value can be chosen)
