question from Sirovich's "Introduction to applied mathematics" This is exercise 20b of Chapter 1 from Sirovich "Introduction to Applied Mathematics": The page with this exercise can also be found here on Google Books:
https://books.google.com/books?id=yOrlBwAAQBAJ&pg=PA10&lpg=PA10

Show that for the harmonic oscillator equation
  $$\frac{dx^2}{dt^2}=\omega^2 x$$
  $w=x+i\omega \frac{dx}{dt}$ satisfies a single first order differential equation.

As far as I can tell, the $w=x+i\omega dx/dt$ satisifes the second-order equation $dw^2/dt^2=\omega^2w$. 
How can I find the first order equation that $w$ would satisfy?
 A: Let's just go by the given equation:
$$\frac{dx^2}{dt^2}=\omega^2 x$$
If $\omega \in \mathbb{R}$ then this is not exactly a harmonic oscillator equation. It has the general solution in terms of hyperbolic functions, or real exponentials:
$$x(t)=A e^{\omega t}+Be^{-\omega t}$$
Writing down the suggested function explicitly:
$$w=A e^{\omega t}+Be^{-\omega t}+i \omega^2 \left(A e^{\omega t}-Be^{-\omega t} \right)=A(1+i \omega^2)e^{\omega t}+B(1-i \omega^2)e^{-\omega t}$$
Which, I believe is not a solution to any first order ODE.

Let's try a harmonic oscillator equation instead:
$$\frac{dx^2}{dt^2}=-\omega^2 x$$
$$x(t)=A e^{i\omega t}+Be^{-i\omega t}$$
$$w=A e^{i\omega t}+Be^{-i\omega t}- \omega^2 \left(A e^{i\omega t}-Be^{-i\omega t} \right)=A(1- \omega^2)e^{i\omega t}+B(1+ \omega^2)e^{-i\omega t}$$
That still is not a solution to a first order ODE.

Now let's try another function instead:
$$u=x+\frac{i}{\omega} \frac{dx}{dt}$$
Now for the HO equation we have:
$$u=A e^{i\omega t}+Be^{-i\omega t}- \left(A e^{i\omega t}-Be^{-i\omega t} \right)=2Be^{-i\omega t}$$
Which is now a solution to a first order equation:
$$\frac{du}{dt}=-i \omega u$$


My only conclusion: there must be a typo in the book (or several).

