$\frac{1}{D+1} e^x$

How do I evaluate $$\dfrac{1}{D+1} e^x$$ where $$D$$ is the differential operator?

I have tried using series expansion, but it just doesn't seem right to me: $$\sum^{\infty}_{k=0}(-D)^ke^x$$

• How should one interpret this notation? Does $\frac1{D+1}$ mean $(D+I)^{-1}$? – Luke Collins Sep 8 at 13:16
• $e^x$ is an eigenfunction of $D$ with eigenvalue $1$. Any sensible interpretation of the expression $\frac{1}{D+1}e^x$ will give you $\frac{1}{2}e^x$. – achille hui Sep 8 at 13:17
• @achillehui Plus the kernel $ce^{-x}$. – Luke Collins Sep 8 at 13:24

If the notation is to be interpreted as $$(D+I)^{-1}$$, then you are looking for function(s) $$y$$ such that $$(D+I)y=e^x.$$ This is equivalent to solving the differential equation $$\frac{dy}{dx}+y=e^x.$$

Multiplying throughout by $$e^x$$, we get $$e^x\,\frac{dy}{dx}+e^xy=e^{2x}\implies \frac{d}{dx}(ye^x)=e^{2x},$$ and integrating both sides yields the general solution $$\boxed{y(x) = \tfrac12e^x + ce^{-x}}$$, where $$c$$ is the constant of integration.

Consider the equation : $$(D-a)y = f (x)$$

i.e. $$(Dy - ay ) = f(x)$$

Now multiply both sides by $$e^{-ax}$$ ( integrating factor )

You can see that :

$$(e^{-ax}Dy -e^{-ax}ay ) = f(x)e^{-ax}$$

Which indeed reduces to :

$$D(e^{-ax} y) = e^{-ax}f(x)$$

Integrate on both sides we get:

$$(D-a) ^{(-1)}f (x) = e^{ax}\int e^{-ax} f(x) dx.$$

Put $$a= -1$$

Now the integral becomes

$$e^{-x}\int e^{x}e^{x}dx$$

Which indeed equals to $$(1/2) e^x$$ ( without considering constant of integration .

• It is not really obvious to me, can you please explain why this property holds? – Max Wong Sep 8 at 14:49
• Sorry, could you please provide a proof for the formula included in the Hint section? – Max Wong Sep 8 at 14:58
• Consider the answer by @Luis Collins . substitute $e^{ax}\int e^{-ax} f(x) dx$ for $y$ see the result. – Alexander supertramp Sep 8 at 14:59
• See the edit @ Max Wong – Alexander supertramp Sep 8 at 15:14
• Thank you. I understand it now. – Max Wong Sep 8 at 15:15