Solving $f(x)= f''(x)$

Question

I am studying limits, continuity and derivability and I have got a continuous function $f\colon \mathbb R \to \mathbb R$ such that $f''(x)= f(x)$. How do I find $f(x)$ from this information? This is the first time I have come across such a problem, that's the reason I am stuck.

I know that its obvious that it should be $e^x$, but what's the formal method of finding the function?

Trigonometric functions like sin or cos don't satisfy the relation.

Wolfram alpha gives the solution to be $f(x)= c_1e^x + c_2 e^{-x}$

Answer 1

You have the second order differential equation ;

$y'' = y \implies y''-y = 0$

The characteristic of the above is

$r^2-1 = 0 $

$r = \pm\sqrt1$

so the two roots are $r=1,-1$

Therefore the two solutions are ;

$y_1(x) = e^x$ and $y_2(x) = e^{-x}$

and now by the principle of superposition you can combine them to get a more general solution;

$y= c_1e^x+c_2e^{-x}$ $\qquad\qquad$ where $c_1,c_2$ are some real constants

Answer 2

There are infinitely many functions that will satisfy $f(x)=f''(x)$ as the function can have some arbitrary constant coefficient. However if you don't take that into consideration, then the most basic and intuitive functions are $e^x$ and $e^{-x}$. As well as any other functions defined using those three functions, including $-e^{-x}$, $\sinh(x)$, and $\cosh(x)$.

Answer 3

If you want a formal way of looking at this, we can use the methods in The Integrator's answer, but justify them. Often, the question of necessity is skipped in explaining this method of solving, i.e. why a solution to the equation must necessarily be of this form. One way to fill in this hole is using the existence-uniqueness theorem, but there is a more direct way of doing it.

Consider (for inspiration), the characteristic polynomial: $$r^2 - 1 = (r - 1)(r + 1).$$ From the $r + 1$ factor, I suggest letting $y(x) = f'(x) + f(x)$ (we could also set $y(x) = f'(x) - f(x)$ and get good results). Then, $$f''(x) - f(x) = f''(x) + f'(x) - f'(x) - f(x) = y' - y = 0.$$ We can now solve $y' - y = 0$, using integrating factors. Multiply through by $e^{-x}$, and we get $$e^{-x} y' - e^{-x} y = (e^{-x} y)' = 0.$$ Therefore, $$e^{-x}y = C \implies y = C e^x$$ for some constant $C$.

Now, we have $$f'(x) + f(x) = y = Ce^x.$$ Using an integrating factor of $e^x$, we get $$(e^x f(x))' = e^x f'(x) + e^x f(x) = Ce^{2x}$$ Integrating both sides, $$e^x f(x) = \frac{C}{2} e^{2x} + D \implies f(x) = \frac{C}{2} e^x + De^{-x}.$$ That is, every solution must satisfy the expected form.

Answer 4

By assuming bijectivity of the function we can deduce the e-function as sole solution: (Let $u$ denote the inverse function of $f$) $$ f''(x) = f(x) \Leftrightarrow f''(u(x)) = x \Leftrightarrow^1 -\frac{u''(x)}{u'(x)^3} = x \Leftrightarrow^2 \frac {1}{u'(x)^2} = x^2 \\\Leftrightarrow u'(x)^2\cdot x^2 = 1 \Leftrightarrow u'(x) = \pm \frac 1 x $$

1: This identity holds for bijective functions and is obtained by differentiating the rule about the derivation of the inverse.
2: Per integration. As we know that both sides are equal, we can drop the constant.

If we now integrate $u'$ and deduct the inverse, we get the two solutions $ e^x, e^{-x}$