# 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

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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