# Method of finding solution of $y'=y+e^x$

(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 $$(D^r + a_1D^{r-1} + \cdots + a_r\mathbf{1})y=f(x)$$?

• Writing the inverse of $D-1$ as expansion series of differential operators is not reasonable to me. Jul 29, 2021 at 10:31
• 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! Jul 29, 2021 at 10:32
• 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. Jul 29, 2021 at 10:33
• 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... Jul 29, 2021 at 10:44
• @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. Aug 1, 2021 at 14:00

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