Solving this DE

Question

I have an ODE as follows-
$ifrac{dU}{dz} = -frac{1}{2}betaomega^2U$
where $U = U(z,omega)$
Separating the variables, I got-
$frac{dU}{U} = -frac{betaomega^2dz}{2i}$
Integrating, I get-
$ln(U) = frac{ibetaomega^2z}{2}$
and hence,
$U = e^{frac{ibetaomega^2z}{2}} + C$.
However, the book says the solution is-
$U(z,omega) = U(0,omega)e^{frac{ibetaomega^2z}{2}}$
What am I doing wrong?

Robert Lewis · Answer

If one integrates the separated equation
$dfrac{dU}{U} = -dfrac{beta omega^2 dz}{2i} tag 1$
'twixt $0$ and $z$, one obtains
$ln U(z,omega) - ln U(0, omega)$
$= displaystyle int_0^z dfrac{dU}{U} = -int_0^z dfrac{beta omega^2 ds}{2i} = idfrac{beta omega^2}{2} int_0^z ds = idfrac{beta omega^2 z}{2} , tag 2$
that is,
$ln left( dfrac{U(z, omega)}{U(0, omega)} right ) = idfrac{beta omega^2}{2} z, tag 3$
whence
$dfrac{U(z, omega)}{U(0, omega)} = exp left (idfrac{beta omega^2 z}{2} right), tag 4$
or
$U(z, omega) = U(0, omega) exp left ( idfrac{beta omega^2 z}{2} right ), tag 5$
in accord with "the book".
What am I doing wrong?  The limits of the definite integral
$displaystyle int_0^z dfrac{dU}{U} tag 6$
we're apparently neglected in our OP Paddy's calculations.

Solving this DE

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