Solving $\frac{dy}{dx}=3\delta\left(x\right)+y\left(x\right)$

Question

The differential equation is $\frac{dy}{dx}=3\delta\left(x\right)+y\left(x\right)$. By separation of variables, we have $\frac{dy}{3\delta\left(x\right)+y\left(x\right)}=dx$. Since $\delta\left(x\right)$ is not constant, the left side cannot be equal to $\ln\left(3\delta\left(x\right)+y\left(x\right)\right)$. It seems that $\delta\left(x\right)$ would be better were it on the right side, since $\int\delta\left(x\right)dx=U\left(x\right)$, the unit step function. However, we cannot separate it from $y$. How should we proceed?

Answer 1

Assuming $\delta(x)$ is the Dirac delta "function", $\delta(x)=0$ for $x \neq 0$ means you can solve the equation over $(-\infty, 0)$ and $(0,\infty)$ while ignoring it. You get a different constant of integration on each interval, so have $$y(x)= \begin {cases} Ae^x& x \lt 0\\Be^x & x \gt 0 \end {cases}$$ Now if you integrate with respect to $x$ over a small interval around $x=0$ we can see that $y(0^+)-y(0^-)=3$ so $B=A+3$. Without an initial condition we cannot find either of $A,B$ explicitly.

Answer 2

$$\begin{align}\frac{dy}{dx}-y&=3\delta(x)\\ u(x)&=e^{ \int-\mathrm{d}x }=e^{-x}\\ e^{-x}\frac{dy}{dx}-e^{-x}y&=3\delta(x)e^{-x}\\ (e^{-x}y)'&= 3\delta(x)e^{-x}\tag{Product Rule}\\ \end{align}$$ Integrate both sides, solve for y, and you’re done.

As it says in your comments, this is a First Order Linear Ordinary Differential Equation. Check out that link for more info.