Solve $y^{\text{'}\text{'}}+(1-i)y^{\text{'}}-iy=0$. Hint: See Chapter 2, Section 10, for a method of finding the square root of a complex number.

Given differential equation is $y^{\text{'}\text{'}}+(1-i)y^{\text{'}}-iy=0$.

To calculate roots of the equation:

$\mathit{x}\mathbf{=}\frac{\left({\displaystyle -b\pm \sqrt{\left(b^2-4\mathrm{ac}\right)}}\right)}{{\displaystyle \mathbf{2}\mathbf{a}}}$

The auxiliary equation is$D^2+(1+2i)D+i-1=0$

According to the question, the roots of this equation is

$$\begin{gathered} D=\frac{-(1+2i)\pm \sqrt{{(1+2i)}^2-4\left(1\right)(i-1)}}{2} \\ D=\frac{-(1+2i)\pm 1}{2} \end{gathered}$$

Any complex number's root could be written as:

$$\begin{gathered} z^{1/2}={\left(r{e}^{i\theta }\right)}^{1/2} \\ =\sqrt{r}\left(\cos \frac{\theta }{n}+i\sin \frac{\theta }{n}\right), \end{gathered}$$

The polar coordinates for$2i$are$r=2$and$\theta =\pi /2$, therefore,

$$\begin{gathered} {\left(2e^{i\theta }\right)}^{1/2}=\sqrt{2}\left(\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}\right) \\ =1+i \end{gathered}$$

Therefore, the auxiliary roots are$D=\frac{-(1-i)\pm (1+i)}{2}$

The equation can be written as

$(D+1)(D+i)y=0$

Solve the simpler equations to obtain two independent solutions, that is

$$\begin{gathered} \frac{dy}{dx}=-y \\ \ln y=-x+c_1 \\ y=c_{2}x{e}^{-x} \end{gathered}$$

And,

$$\begin{gathered} \frac{dy}{dx}=iy \\ \ln y=ix+c_2 \\ y=c_4{e}^{ix} \end{gathered}$$

Therefore, the general solution is$y=c_{2}x{e}^{-x}+c_4{e}^{ix}$

It shows a different form of the solution that it gets when use the methods of solution for a real roots or complex conjugates roots.