Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$ I need help with this exercise.
Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$
a) Let $g(x)=f(x)e^{-x}$. Find $g'(x)$.
So this is what I did
$$g'(x)=f'(x)\cdot e^{-x}+e^{-x}\cdot -1 \cdot f(x)$$
$$g'(x)=f'(x)\cdot e^{-x}-e^{-x} \cdot f(x)$$
$$g'(x)=e^{-x}[f'(x)-f(x)]$$
$$g'(x)=e^{-x}[f(x)+1-f(x)]$$
$$g'(x)=e^{-x}[1]$$
$$g'(x)=e^{-x}$$
b) Find all the differentiable functions $f$ satisfying $f'(x)=f(x)+1$
I'm stuck with this one. Please help!
 A: The second part just needs you to solve the differential equation. If you let $f(x) =y$, you can write $\frac {dy} {dx} = y+1$.
You can separate and integrate to get: $\int \frac 1{y+1}dy = \int 1 dx$.
You then get: $\ln|y+1| = x + c$, where $c$ is an arbitrary constant.
Taking the exponent of both sides:
$|y+1| = e^xe^c = Ae^x$ (we're just assigning another constant $A>0$ such that $A = e^c$).
Now, since the right hand side is always positive, we can remove the absolute value sign and rearrange to get $y = \pm(Ae^x) - 1$, which you can write as: $f(x) =Be^x - 1, \ B \ne 0$.
You now note that when $B =0$, you get $f(x) = - 1$ which trivially satisfies the differential equation.
Hence the final solution is $f(x) = Be^x - 1$, without restriction on $B$. If you're working in the reals, $B$ can be any real number.
A: $$g'(x)=e^{-x}$$
$$g(x)=-e^{-x}+C$$
$$f(x)=g(x)e^x=-1+Ce^x$$
A: You should really have explained in your question that you needed to use (a) to solve (b).
Spoilers ahead: click each one for a hint. (In the following, DE stands for Differential Equation)

 Can you rewrite the DE for f using what you got in (a)?


 $f(x)= g(x)e^x$, so what is f' using the product rule?


 Anyway, have you learned about "integrating factor", or "exact differentials"?

