# 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?

• If $y=c e^{x}$ and $y=ke^{-x}$ are solutions then $y=ce^{x}+ke^{-x}$ is a solution . – randomgirl Apr 28 '15 at 16:56
• The general solution is $f(x) = c_1e^x + c_2e^{-x}$ for arbitrary real constants $c_1, c_2$. Proving these are the only solutions is usually dealt with by using an atom bomb strength theorem about the existence & uniqueness of solutions of ODEs. – Simon S Apr 28 '15 at 16:56
• @Simon Atom bombs are kewl! And I like theorems! – bjb568 Apr 28 '15 at 16:58
• For completeness: the general solution can also be written as $f(x)=c_1\cosh x+c_2\sinh x$, using Hyperbolic trigonometric functions. This is more convenient for finding $f$ with given initial conditions: e.g., if one wants $f(0)=5$ and $f'(0)=-2$, the solution is $5\cosh x-2\sinh x$. – user147263 Jun 5 '15 at 18:44

Non atom-bomb proof that $A e^x + B e^{-x}$ are the only solutions:

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

2. $(y'-y)' = y'' - y' = y - y' = -(y'-y)$ 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.

• well done. nice work. – abel Apr 28 '15 at 19:48
• It's good to know how a skillful karate chop or two will do the trick. ;-) – Simon S Apr 29 '15 at 13:37

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.

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.

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 $D$.