This 2nd order ordinary differential equation

$$a f''(x) + k f(x) = 0, \ a, k \in \mathbb{R}$$

has the following characteristic polynomial:

$$a y^2 + k = 0$$

whose roots are

$$y = \pm i \sqrt{\frac{k}{a}}$$

So two solutions are:

$$f_1 (x) = e^{i \sqrt{\frac{k}{a}} x}, \ f_2 (x) = e^{-i \sqrt{\frac{k}{a}} x}$$

A linear combination of $f_1$ and $f_2$, $f = C_1 f_1(x) + C_2 f_2 (x)$ is still a valid solution.

1) Why? Because the derivative is a linear operator and the initial equation is linear, too?

2) Is $f$ still a solution if $C_1, C_2 \in \mathbb{C}$? If yes, why? Is it still due to the linearity of derivative and the linearity of the equation itself, which doesn't change when dealing with complex numbers?

For example, to obtain $g_1 (x) = \cos( \sqrt{k / a} x)$ and $g_2 (x) = \sin( \sqrt{k / a} x)$ as solutions, it must be $C_1, C_2 \in \mathbb{C}$.


1 Answer 1


1) Yes, for the reasons you mentioned, but it is also very important that the right hand side of the differential equation be 0 for this property to hold.

2) Yes. You are right that the linearity of the derivative does not care about complex numbers in this case. Note that you never specified whether the solutions you are were looking for should be real or complex valued. If you specified that you functions had to be real-valued, then you would have to impose some other restrictions on $C_1$ and $C_2$.


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