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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 ?

Salutsalut1
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2 Answers2

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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 "$r_1=r_2$".)

David C. Ullrich
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  • See also here: https://math.stackexchange.com/questions/2625995/on-the-solutions-to-an-n-th-order-linear-homogeneous-constant-coefficient-ode/2625996#2625996 – fred goodman Jan 29 '18 at 03:12
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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.

fred goodman
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    $\sinh(x)$ and $\cosh(x)$ is another "canonical" pair of functions that spans the space. – Ben Grossmann Jan 26 '18 at 15:53
  • But this isn't really a proof, because the typical beginning book on differential doesn't prove that the solution space has dimension $2$. – David C. Ullrich Jan 26 '18 at 17:05
  • This is an interesting comment. I had a look at a recent edition of Boyce and Diprima, which I take to be typical, or at least mainstream. It does contain a proof that the solution space is 2 dimensional modulo assuming without proof the existence and uniqueness theorem for the IVP. It also has a discussion of the Wronskian, which perhaps just muddies the waters. I think your contribution is to show that it is quite unnecessary to appeal to the uniqueness theorem for the IVP in this context. – fred goodman Jan 26 '18 at 20:05
  • Of course it follows from existence and uniqueness - not that you're implying otherwise, but that's a pretty big gap. Yes, the Wronskian stuff in B&P strikes me as a little silly, given that the actual method we're going to learn to identify the basis functions is to follow the algorithm... – David C. Ullrich Jan 27 '18 at 00:35