Solving $v'(x) - ie^{x/2} v(x) = e^{2ie^{x/2}}$

Question

While considering the first order inhomogenous ODE $$ v'(x) - ie^{x/2} v(x) = e^{2ie^{x/2}}, $$ I became curious about the most concise way to produce a solution. For the homogenous equation $$v_h'(x)-ie^{x/2}v_h(x) = 0$$ that I will use to satisfy the boundary conditions, I employed a separation of variables to determine $v_h(x) = Ce^{2ie^{x/2}}$. For a particular solution $v_p(x)$ of the inhomogeneous equation, I had thought about employing a method of integrating factors, though I noticed such a method worked well only for particular coefficients. I feel as though a variation of parameters or Laplace transform would be far too much for the problem. Does anyone have simple method for solving the above equation?

Thank you all.

Answer 1

We take $$ \frac{d}{dx} \left(e^{-2ie^{x/2}}v(x)\right) = v'(x)e^{-2ie^{x/2}} -i e^{-2ie^{x/2}} v(x)=e^{2ie^{x/2}}. $$ Although the resulting formula is not particular nice, we conclude as follows.

$$v_p(x) = \exp\left(2ie^{x/2}\right) \int_0^x \exp\left(2i e^{s/2}\right) ds$$

For a method to evaluate this integral, please refer to the following question.

Answer 2

For the homogeneous equation, you found $$v=C \,e^{2 i e^{x/2}}$$ Variation of the parameter gives $$C'=1 \implies C=x + k \implies v= ???$$