Method to solve a first order ODE with convolution exponential

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In thermodynamics, I have an equation like this in frequency domain $\omega$ and spatial domain $z$:

$$ \partial_z g(z,\omega) - f(\omega) \ast g(z,\omega) = h(z,\omega) $$

with the boundary condition: $g(0,\omega) = 0$. The functions $h$, $f$ and $g$ are complex function with any complex value. The convolution product is in frequency domain:

$$ f_1(\omega) \ast f_2(\omega) = f_3(\omega) $$

What is the procedure to solve this ODE (in frequency or time domain with an inverse Fourier transform)?

What I did but I am not sure

I tried something that seems working but I would like to know if it is possible and if it can be extended to higher order ODE? So I rewrite the equation as:

$$ \left( \partial_z - f(\omega) \ast \right) g(z,\omega) = h(z,\omega) $$

Then directly I state that:

$$ g = e^{z f(\omega) \ast} C(z,\omega) $$

with:

$$ e^{z f(\omega) \ast} C \simeq C + z f(\omega) \ast C + \frac{1}{2} z^2 f \ast (f \ast C) + ... $$

and $C$ a function to be determined. Then I have the ODE for $C$:

$$ e^{z f(\omega) \ast} \partial_z C = h(z,\omega)$$

Then I "inverse" the exponential convolution (do not know if I can brutally like this):

$$ \partial_z C = e^{- z f(\omega) \ast} h(z,\omega) $$

I integrate and obtain (I know from boundary condition that $g(0,\omega) = 0$):

$$ C = \int_0^z e^{- z' f(\omega) \ast} h(z',\omega) d z' $$

So at the end we have the solution:

$$ g = e^{z f(\omega) \ast} \int_0^z e^{- z' f(\omega) \ast} h(z',\omega) d z' $$

Is it correct ?