If the characteristic equation for a differential equation can be written as $(s-r_1)(s-r_2)$, the substition $z=y'-r_1y$ yields an equation of the form $z'-r_2z=f(x)$.

For example, if our equation is

$y''-3y'+2y=e^x$,

the substitution

$z=y'-2y$

simplifies to

$y''-2y'-(y'-2y)=z'-z=e^x$

At this point, integrating factors can be used to solve for $z$, then substituting back will yield a solution for $y$.  My first question is why does it work like this?

My second question is if there's a way to find a substitution for a general second order linear ODE

$y''+p(x)y'+q(x)y=f(x)$

that will similarly reduce the problem to a first order linear ODE which can then be solved by integrating factors?

Define $g(z)=\overline{f(\overline z)}$.  Then note that $f(z)=\overline{g(\overline z)}$, so there is symmetry allowing only one implication to be shown directly.  Suppose that $f$ is analytic.  Then for all $z$, 

$$\begin{align*}g'(z)&=\lim\limits_{h\to 0}\frac{g(z+h)-g(z)}{h}\\
&=\lim\limits_{h\to 0}\frac{\overline{f(\overline{z+h})}-\overline{f(\overline z)}}{h}\\
&=\lim\limits_{h\to 0}\overline{\left( \frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right)    }\\
&=\overline{\left( \lim\limits_{h\to 0}\frac{f(\overline z+\overline h)-f(\overline z)}{\overline h} \right)    }\\
&=\overline{f'(\overline z)}.
\end{align*}$$

That is, $g$ is differentiable, with $g'(z)=\overline{f'(\overline z)}$.  



Alternatively, as other answers have indicated, you could check that the Cauchy-Riemann equations hold for $g$ if they hold for $f$, with $f(x+iy)=u(x,y)+iv(x,y)$ and $g(x+iy)=u(x,-y)+i(-v(x,-y))$ as you indicated.



Another perhaps more conceptual way to think of this is that complex analytic maps are conformal (where their derivatives are nonzero), preserving orientation and angles.  Complex conjugation preserves angles but reverses orientation.  Reversing orientation twice gets you back where you started, so the result is that $g$ is conformal.  (I have given an idea here rather than anything close to a rigorous proof.)



Another approach is to look at power series expansions.  If $f$ has power series expansion in a neighborhood of $\overline{c}$, $\displaystyle{f(z)=\sum\limits_{k=0}^\infty a_k(z-\overline c)^k}$, then in a neighborhood of $c$, $g$ has the power series expansion $\displaystyle{g(z)=\sum\limits_{k=0}^\infty\overline{a_k}(z-c)^k}$.  That is, you just conjugate the coefficients and conjugate the base point for the expansion.  This shows that $g$ is analytic if $f$ is.
