# What allows us to use the Heaviside operator like a variable?

We were taught to use the Heaviside operator $$D: \dfrac{d}{dx}$$ to solve an ODE, for example,

Consider $$y'' + 3y' +2y = e^{-2x}$$

$$\implies (D^2 + 3D + 2)y = e^{-2x}$$ $$\implies y = \dfrac{1}{D^2 + 3D + 2} e^{-2x}$$ $$\implies y = \dfrac{1}{(D+1)(D+2)} e^{-2x}$$ Now we substitute $$-2$$ in place of $$D$$, in this case the denominator becomes zero, so we differentiate the denominator with respect to $$D$$ as if it's a variable and multiply$$^*$$ a factor $$x$$.

$$y = \dfrac{x}{(D+1)+(D+2)} e^{-2x}$$ And then do the substitution $$\implies y = -xe^{-2x}$$

How is this possible, how are we able to treat an operator $$\dfrac{d}{dx}$$ like a variable? What does differentiate with respect to $$D$$ even mean?

$$*:$$ I also don't understand why we multiply $$x$$ in the numerator

• That is horrible - my sympathies. This is possibly a symbolic representation of the Laplace transform method to find a particular solution of the equation. There differentiation is replaced by multiplication by a variable, which you can then differentiate with respect to.
– Paul
Commented Nov 21, 2019 at 12:26
• @Paul Sorry I don't get it, what's horrible? I hope Laplace transform and this method are different, I've written an answer that's somewhat related to this(math.stackexchange.com/a/3430967/525644), that led me to another question(math.stackexchange.com/questions/3431062/…) which finally led me to ask this one. Commented Nov 21, 2019 at 12:30
• Treating an operator exactly like a variable is pretty horrible to start with. As you said, what does differentiate with respect to D even mean.
– Paul
Commented Nov 21, 2019 at 12:33
• @Paul Unfortunately this is how we are taught:( Please take a look at the other question, it would be really helpful if you can answer :) Commented Nov 21, 2019 at 12:35
• I have time: $$\int_0^{\infty} \left(xe^{-ax}\right)e^{-sx}dx = \int_0^{\infty} \frac{-x}{a+s} d\left[e^{-(a+s)x}\right] = - \left[\frac{x\,e^{-(a+s)x}}{a+s}\right]_0^{\infty} + \frac{1}{s+a}\int_0^{\infty} e^{-sx}e^{-ax} dx$$ The latter integral is in your list, so Laplace$\left(xe^{-2x}\right) = 1/(s+2)^2$. Commented Nov 26, 2019 at 18:53

In the QM version of Operator Calculus as I learned it at the university, a formula is derived that should be stored in non-volatile memory. It's in the box near the bottom of our general theory: $$\large \boxed{\; \frac{d}{dx} + f = e^{-\int f \, dx}\, \frac{d}{dx}\, e^{+\int f \, dx } \;}$$ Strangely enough, you have actually derived part of this memorable formula yourself, in this answer, where you write: $$D[e^{at}\,y(t)]=e^{at}(D+a)[y(t)]$$ From which the equivalent follows: $$\left[\frac{d}{dt}+a\right]y(t) = \left[e^{-at}\frac{d}{dt}e^{+at}\right]y(t)$$ Herewith your problem can be solved. Note that there are no fractions involved with $$D=d/dx$$ in the denominator. $$\left[\left(\frac{d}{dx}\right)^2+3\left(\frac{d}{dx}\right)+2\right]y=e^{-2x}\\ \left[\frac{d}{dx}+1\right]\left[\frac{d}{dx}+2\right]y=e^{-2x}\\ \left[e^{-x}\frac{d}{dx}e^{+x}\right]\left[e^{-2x}\frac{d}{dx}e^{+2x}\right]y=e^{-2x}\\ \frac{d}{dx}e^{+x}.e^{-2x}\frac{d}{dx}e^{+2x}y=e^{-x}\\ e^{-x}\frac{d}{dx}e^{+2x}y=\int e^{-x}dx = -e^{-x}+C_1\\ \frac{d}{dx}e^{+2x}y=1+C_1e^{x} \quad \mbox{(!)}\\ e^{2x}y=\int \left[ 1+C_1e^x \right] dx = x + C_1 e^x + C_2\\ y = x.e^{-2x} + C_1 e^{-x} + C_2 e^{-2x}$$ Where the constants $$C_1$$ and $$C_2$$ are still to be determined from initial or boundary conditions.
• No, instead I'd rather discourage the "differentiating" thing you mentioned. Please take a look at the general theory I mentioned earlier. Then you will see that doing algebraic calculations with your $D$ is the same (most of the time) as working with common (complex) numbers. Provided, though - and this is important - that there are only constant coefficients involved with the differential equation at hand. If otherwise, then there possibly is an issue with operators being non-commutative, which is quite another challenge. Commented Nov 26, 2019 at 18:21