(Edited after serious comments from Chongxu Ren, Ulrich, Gerry Myerson etc; thanks to them for bringing my attention to put the question in precise form)

I am trying to find the solution of differential equation $y'=y+e^x$ for $x\in\mathbb{R}$.

This can be solved by using method of integrating factor; but without referring to it, I went for looking solution step-by-step with complexity, as follows.

Simple Case: If it would have been $y'=e^x$, i.e. $Dy=e^x$, we could have taken integration of both sides to get $y=ae^x+b$.

(The method of integrating factor brings above equation to form $D(e^{-x}y)=1$)

General Case: The right side is not only the function of $x$, but some terms of the (unknown) function $y$ also; an easy example is the equation in title.

We then move to collect terms of $y$ on one side, and keep function of $x$ other side. We can write main equation as $$(D-\mathbf{1})y=e^x,$$ and we want to invert $D-\mathbf{1}$ to get information of unknown $y$. [In case $Dy=f(x)$, we expect to get $y=\int f(x)$, provided, $f(x)$ satisfies some conditions on given domain of it.]

For $(D-\mathbf{1})y=e^x$, I went to do like: $y=\frac{1}{D-\mathbf{1}}e^x= (\mathbf{1}+D+D^2+\cdots )(-e^x)$. But, this last expression on RHS does not make sense since $D^n(-e^x)=-e^x$ for all integers $n\ge 0$.

Question: When does such method of inverting $D-\mathbf{1}$ or a polynomial expression of $D$ actually works to give solution of given differential equation - say $(D^r + a_1D^{r-1} + \cdots + a_r\mathbf{1})y=f(x)$?

  • $\begingroup$ Writing the inverse of $D-1$ as expansion series of differential operators is not reasonable to me. $\endgroup$
    – Hgtcl
    Jul 29, 2021 at 10:31
  • $\begingroup$ If $1+D+D^2+\cdots$ is going to work at all, it should work when the thing on the right is a polynomial in $x$, say, $x$ itself. Try it, see if it works! $\endgroup$ Jul 29, 2021 at 10:32
  • $\begingroup$ If you think $D-1$ as functional operator between smooth function space, you'll find that one can't apply all arithmetic rule to it. $\endgroup$
    – Hgtcl
    Jul 29, 2021 at 10:33
  • 4
    $\begingroup$ exactly how do we get from $y'=y$ to $y=ce^x$ just by integrating both sides??? I believe that $\int y'=y$, the problem is the RHS... $\endgroup$ Jul 29, 2021 at 10:44
  • 1
    $\begingroup$ @GerryMyerson: (D-1)y=x so right side is polynomial in x, and so, if we write $y=\frac{-1}{1-D}x=(-1-D-D^2-\cdots )x=-x-1$, so $y=-1-x$ is solution by this method, which is in fact a solution. $\endgroup$ Aug 1, 2021 at 14:00

1 Answer 1


Here is way to make the integrating factor method work in this specific case. Multiplying by $e^{-x}$, the equation is equivalent to $$y'e^{-x}-ye^{-x}=1 \quad \quad \quad \quad \text{ i.e. to } \quad \quad \quad \quad(ye^{-x})'=1$$

So you get $$ye^{-x}=x+C\quad \quad \quad \quad \text{ i.e. } \quad \quad \quad y=(x+C)e^x$$

  • 9
    $\begingroup$ While this is a solution method but this is not an answer to OP's question. OP has already mentioned the integrating factor method at the beginning. $\endgroup$
    – Anurag A
    Jul 29, 2021 at 10:14
  • $\begingroup$ @AnuragA I agree with you, but I thought the OP would be happy to see how to make the method work in that precise situation, even if it does not answer his/her last question. $\endgroup$ Jul 29, 2021 at 10:29

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