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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

Reference: https://www.youtube.com/playlist?list=PLEC88901EBADDD980

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Aravindh Vasu
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    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 Nov 21 '19 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(https://math.stackexchange.com/a/3430967/525644), that led me to another question(https://math.stackexchange.com/questions/3431062/how-is-laplace-transform-more-efficient?noredirect=1#comment7079765_3431062) which finally led me to ask this one. – Aravindh Vasu Nov 21 '19 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 Nov 21 '19 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 :) – Aravindh Vasu Nov 21 '19 at 12:35
  • Ah, you seem to have a perfectly good understanding of what you are talking about. – Paul Nov 21 '19 at 12:40
  • @Paul Nah, in that answer I linked, once I arrive at an expression $y= H(D)x$ I use partial fractions and separate H(D) into two first order equations or just cascade them without doing partial fraction expansion. But I'm not able to really relate these with the method mentioned above. The thing bugs me is that factor x, where did that even come from. – Aravindh Vasu Nov 21 '19 at 12:45
  • In the partial fraction generated by Laplace transforms you get a double root $\frac{1}{(s+2)^2}$ in the denominator, which essentially comes from a derivative of $e^{-sx}$ with respect to s. If the right hand side was $e^{-3x}$ there would be no double root and no multiplication by x I guess. – Paul Nov 21 '19 at 12:50
  • @Paul Can you please elucidate a bit? – Aravindh Vasu Nov 21 '19 at 12:52
  • Sorry, no time. – Paul Nov 21 '19 at 13:56
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    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$\left(xe^{-2x}\right) = 1/(s+2)^2$.

     – Han de Bruijn Nov 26 '19 at 18:53

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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.

Han de Bruijn
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  • Okay, so that "differentiating" thing I mentioned is just a trick to remember this? But we also do partial fraction expansion and all kinds of "polynomial-stuff" with the operators, what justifies that? What I'm expecting is that; Do we treat polynomials of D like a normal polynomial just because it works, or we have some deeper reason as to why that works? – Aravindh Vasu Nov 26 '19 at 15:48
  • 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. – Han de Bruijn Nov 26 '19 at 18:21