This doesn’t completely explain the trick, though. In the step where you write f=(1-I)^{-1} 0, if you interpret I as an operator then you get f=0 as the result. To get f=Ce^x you need to have f=(1-I)^{-1} C in that step instead. You can get this by replacing \int f by If+C at the beginning.
Ah sorry, I skipped over that derivation! Here’s how we’d approach this from first principals: to solve f=Df, we know we want to use the (1-x)=1+x+x^2+… trick, but now know that we need x=I instead of x=D. So that’s why we want to switch to an integral equation, and we get f=Df If=IDf = f-f(0) where the final equality is the fundamental theorem of calculus. Then we rearrange: f-If=f(0) (1-I)f=f(0) and solve from there using the (1-I)=1+I+I^2+… trick! What’s nice about this is it shows exactly how the initial condition of the DE shows up.
This doesn’t completely explain the trick, though. In the step where you write f=(1-I)^{-1} 0, if you interpret I as an operator then you get f=0 as the result. To get f=Ce^x you need to have f=(1-I)^{-1} C in that step instead. You can get this by replacing \int f by If+C at the beginning.
Ah sorry, I skipped over that derivation! Here’s how we’d approach this from first principals: to solve f=Df, we know we want to use the (1-x)=1+x+x^2+… trick, but now know that we need x=I instead of x=D. So that’s why we want to switch to an integral equation, and we get
f=Df
If=IDf = f-f(0)
where the final equality is the fundamental theorem of calculus. Then we rearrange:
f-If=f(0)
(1-I)f=f(0)
and solve from there using the (1-I)=1+I+I^2+… trick! What’s nice about this is it shows exactly how the initial condition of the DE shows up.