# Complex exponential method of solving differential equations

In the twenty third Feynman lecture, the solution of the following differential equation is discussed:

$$\frac{d^2 x}{dt^2} + \frac{kx}{m} = \frac{F}{m}$$

AFter 'complexifying' this differential equation, he gets:

$$\frac{d^2 x}{dt^2} + \frac{kx}{m} = \frac{\hat{F} e^{iwt} }{m}$$

And then it is written that we can write:

$$x = |x| e^{i \omega t}$$

Assuming $$x$$ is a complex number and this leads to:

$$\frac{dx}{dt} = i \omega x$$

However, the above result assumes that $$|x|$$ is constant, how do we rigorously justify this assumption?

• Mar 24, 2021 at 21:00
• I'm not sure understand what you're looking for vis-à-vis a 'rigorous justification'. Isn't it just that one is seeking solutions of such form in order to be able to express a general solution over such a basis? Mar 24, 2021 at 23:33
• Hmm , the concept of solution basis and other's feels outside the text which I have linked. @AlfredCentauri Mar 25, 2021 at 7:08
• No effort then to clarify what it is that you're looking for? Mar 26, 2021 at 2:48
• I mean, how exactly do you think I can clarify the question? If you see the text of Feynman, he explains without bringing up the concepts you've mentioned (however it is unclear to me) Mar 26, 2021 at 5:28

$$x$$ and $$F$$ can each be expressed as a Fourier integral: $$x(t)=\int x(\omega)e^{i\omega t} d\omega$$ $$F(t)=\int F(\omega)e^{i\omega t} d\omega$$ This of course assumes that $$F(t)$$ is square integrable ($$L^2$$ Hilbert space).