I would like to find all $f \in C^2(\mathbb{R})$ such that : $$f'' = f$$
I don't how to procceed so I try the trick of dividing both LHS and RHS by $f'$ in order to regonize some well know derivative ($\ln(f)$) but it doesn't seem to work... Any ideas ?
On
Here's an actual proof, missing from the typical DE book: Say $f''=f$. A little trick, inspired by the identity $D^2-I=(D-I)(D+I)$:
Let $g=f'+f$. Then $g'-g=0$.
Hence $$(e^{-x}g(x))'=0,$$so $e^{-x}g(x)$ is constant: $$g(x)=c_1e^x,$$ or $$f'+f=c_1e^x.$$Repeat the same trick, multiplying by $e^x$:
$$(e^xf(x))'=c_1e^{2x},$$so $$e^xf(x)=\frac12c_1e^{2x}+c_2;$$hence $$f(x)=\frac12c_1e^x+c_2e^{-x}.$$
Exercise Use the same method to find all the solutions to $f''-2f'+f=0.$ (Hint: $D^2-2D+I=(D-I))(D-I)$.)
(That's one of the things I like about doing it this way - it works the same regardless of whether "$r_1=r_2$".)
This is a second order linear homogeneous ODE. The set of solutions is a linear space of dimension 2. You can look up the usual techniques for such an equation, but in this case you can see by inspection that $\{e^x, e^{-x}\}$ is a pair of linearly independent solutions, hence a basis of the solution space.
Edit: The proof that $\{e^x, e^{-x}\}$ is a basis of the solution space given in the text of Boyce and Diprima (which is possibly typical) goes like this. Denote $f_1(x) = e^x$, $f_2(x) = e^{-x}$. Not only are these linearly independent but their vectors of initial values are linearly independent: $$ v_1 = \begin{bmatrix} f_1(0)\\f_1'(0)\end{bmatrix} = \begin{bmatrix} 1 \\1 \end{bmatrix},$$ $$ v_2 = \begin{bmatrix} f_2(0)\\f_2'(0)\end{bmatrix} = \begin{bmatrix} 1 \\-1 \end{bmatrix}.$$ It follows that for any solution $g$ of the DE, there exists a solution $f = c_1 f_1 + c_2 f_2$ which matches initial conditions: $f(0) = g(0)$ and $f'(0) = g'(0)$. An appeal is then made to the uniqeuesss theorem for the initial value problem to conclude $f = g$. The same proof is given in general for $n$th order homogeneous linear constant coefficient DE's.