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

Question

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: $$\sum^{\infty}_{k=0}(-D)^ke^x$$

Answer 1

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.

Answer 2

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 .