Performing Differentiation on Complex Functions!

Question

Using method of variation of parameters , solve the following differential equation $$\frac{d^2y}{dx^2}+y=\csc x.$$

I know that, if, the given differential equation is of the form, $\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=g(x),$ and the two fundamental solution of $\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=0,$ are $u$ and $v$ then, the particular integral of $$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=g(x),$$ is given by $$y=-u\int\frac{vg(x)}{w(u,v)}dx +v\int\frac{ug(x)}{w(u,v)}dx\tag 1,$$ where $w(u,v)$ is the wronskian of the two functions $u,v$ i.e $w(u,v)=\begin{vmatrix} u & v\\ u'&v'\end{vmatrix}.$

However, the problem is, in this case, the complementary function, or the fundamental set of solutions of $\frac{d^2y}{dx^2}+y=0$ are $u=e^{ix}$ and $v=e^{-ix}.$

Here, I am experiencing a couple of issues while applying the above mentioned formula $(1)$ of particular integrals.

All in all, I have to calculate the wronskian firstly. But then, I have only worked with calculus problems in real functions. This is the first time, I am trying to differentiate a complex function as, $u=e^{ix}.$

I am unsure whether, $u'=ie^{ix}.$ I tried emulating the same thing that we do in case of real functions.

Then, comes the next hurdle similar to these. How do we integrate $$y=-u\int\frac{vg(x)}{w(u,v)}dx +v\int\frac{ug(x)}{w(u,v)}dx,$$ when, $u,v$ are complex functions?

Say, for example, $\int\frac{vg(x)}{w(u,v)}=\int\frac{e^{-ix}\csc(x)}{w(e^{ix},e^{-ix})}$ I have no whatsoever idea about this whole process, since never did I perform an integration on complex functions.

Any clarification regarding this apparent issues will be highly appreciated.

Answer 1

Inspired from the comment of @Gonçalo, I solved the problem, avoiding the complex functions.

We first note that if $$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=g(x),$$ is the given differential equation then the corresponding homogeneous differential equation is $$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=0\tag 1$$

If $u,v$ are a fundamental set of solutions to $(1)$ then, $u_1=(u+v)$ and $v_1^*=(u-v)$ are solutions of $(1)$ as well and they will be furthermore, linearly independent.

In this case, a simple calculation indicates $u_1=2\cos x$ and $v_1=\frac{1}{2i}v_1^*=\frac{1}{2i}{2i\sin x}=\sin x.$ We can observe, that $v_1$ is a solution of $(1)$ and linearly independent of $u_1.$

Now, we assume, the fundamental set of solutions to be $u_1=\cos x$ and $v_1=\sin x.$

The wronskian $w(u_1,v_1)=1.$

We use the formula of the method of variation of parameters as follows:

$$y=-u\int\frac{vg(x)}{w(u,v)}dx +v\int\frac{ug(x)}{w(u,v)}dx\implies -\cos x\int \csc x \sin xdx+\sin x\int \cos x\csc xdx=-\cos x\int dx+\sin x\int \frac{\cos x}{\sin x}dx=-x\cos x+\sin x\log |\sin x|.$$

So, the complete solution is, $$y=c_1e^{ix}+c_2e^{-ix}-x\cos x+\sin x\log |\sin x|.$$