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Solving this DE

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?

Mathematics Asked on November 12, 2021

1 Answers

One 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.

Answered by Robert Lewis on November 12, 2021

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