For what functions is $y'' = y$? What functions $y = f(x)$ have the property that $f(x) = f''(x)$, i.e. what functions have the same integral and derivitive?
I could think of $ce^x$ and $ce^{-x}$ (where $c$ is a constant), but are there others? If not, how can you prove those are the only ones?
 A: Multiply both sides by $2y'$ and you get
$$2y''y'=2y'y.$$
You should recognize the derivative of $y'^2$ on the left and that of $y^2$ on the right.
Integrating,
$$y'^2=y^2+C_0,$$
and
$$\frac{y'}{\sqrt{y^2+C_0}}=\pm1.$$
Integrating once again (using a table),
$$\ln\left(y+\sqrt{y^2+C_0}\right)=\pm x+C_1,$$
or
$$y+\sqrt{y^2+C_0}=C_2e^{\pm x},$$
$$y^2+C_0=\left(C_2e^{\pm x}-y\right)^2,$$
$$C_0=C_2^2e^{\pm2x}-2C_2e^{\pm x}y,$$
$$\color{green}{y=Ce^{\pm x}+C'e^{\mp x}}.$$
This confirms the well-known result.
A: All of the linear combinations of the two solutions that you mentioned. 
That is,
$ y = A e^{x} + B e ^{-x} $ where $ A,B $ are constants.
A: Using the differential operator $D$, the equation is
$$D^2y=y\text{, or }(D^2-1)y=0.$$
This factors as 
$$(D-1)(D+1)y=0.$$
We first set $z=(D+1)y$ and solve
$$(D-1)z=0.$$
$$Dz-z=0\implies (Dz)e^{-x}-ze^{-x}=(Dz)e^{-x}+zD(e^{-x})=D(ze^{-x})=0\implies ze^{-x}=C_0,$$
$$z=C_0e^x.$$
Then solve
$$(D+1)y=C_0e^x.$$
$$Dy+y=C_0e^x\implies (Dy)e^x+ye^x=(Dy)e^x+yD(e^x)=D(ye^x)=C_0e^{2x}\\\implies ye^x=C_1e^{2x}+C_2,$$
$$\color{green}{y=Ce^x+C'e^{-x}}.$$
The method generalizes to arbitrary polynomials in $D$.
A: Non atom-bomb proof that $A e^x + B e^{-x}$ are the only solutions:


*

*$(y'+y)' = y'' + y' = y + y'$ so $y'+y = C e^x$

*$(y'-y)' = y'' - y' = y - y' = -(y'-y)$ so $y'-y = D e^{-x}$
Subtracting these two equations we get $2y = C e^x - D e^{-x}$, or, after dividing and renaming constants:
$y = A e^x + B e^{-x}$

Supplement proving if $u' = r u$ then $u = C e^{rx}$:
Rewrite this as $u' - ru = 0$ and multiply both sides by $e^{-rx}$.
Then $e^{-rx} u' + (-r) e^{-rx} u = 0$.
The LHS is $\frac{d}{dx}\left[ e^{-rx} u\right]$, and the equation is stating that this derivative here is zero, so we get:
$e^{-rx} u = C$
or in other words
$u = C e^{rx}$

Background on these two methods:
The trick in the top section is a general technique for proving results for systems of linear first order equations based on finding eigenvectors. (A second order equation is a system of two first order equations if you let $y'$ be a new variable and include $(y)' = y'$ as one of your equations.)
The trick in the second section is called the method of integrating factors.
There's also a super fancy way to do both sections at once using a matrix exponential.
