solve coupled differential equations How do I solve the following coupled linear differential equations
$$
\dot{x}(t)=iAy(t)-iBx(t)\\
\dot{y}(t)=iAx(t)+iAz(t)\\
\dot{z}(t)=iAy(t)-iBz(t)
$$
for $|x(t)|^2, |y(t)|^2, |z(t)|^2$ ?
given the initial conditions $x(0)=1,y(0)=z(0)=0$ and $|x(t)|^2+|y(t)|^2+|z(t)|^2=1$
Note: A, B are constants and $i=\sqrt{-1}$
 A: you can use Laplace Transform as follow to find the answer.(a fast and easy way to know Laplace Transform is through wikipedia https://en.wikipedia.org/wiki/Laplace_transform) by taking Laplace Transform of above equations we get:
$$\dot{x}(t)=iAy(t)-iBx(t)\Rightarrow sX(s)-1=iAY(s)-iBX(s)\\ 
\dot{y}(t)=iAx(t)+iAz(t)\Rightarrow sY(s)= iAX(s)+iA(s)\\
\dot{z}(t)=iAy(t)-iBz(t)\Rightarrow sZ(s)=iAY(s)-iBZ(s)$$
now we can treat equations in s-domain like a linear system of equations with complex variables and also variable $"s"$ as a constant. so we must solve for $X(s),Y(s),Z(s)$  from equation's. 
it is easy to prove that:
$$X(s)=\frac{iAY(s)+1}{s+iB}\\ 
iA-A^2Y(s)-s(s+iB)Y(s)-A^2Y(s)=0\\
Z(s)=\frac{iAY(s)}{s+iB}$$
so after finding value of $Y(s)$ from last equation we plug it into the other two and find values of other functions ($X(s),Z(s)$):
$$X(s)=\frac{s^2+iBs+A^2}{(s^2+iBs+2A^2)(s+iB)}\\ 
Y(s)=\frac{iA}{s^2+iBs+2A^2}\\
Z(s)=\frac{-A^2}{(s^2+iBs+2A^2)(s+iB)}$$
now we must take Inverse Laplace Transform from above equations using Partial Fraction Expansion but this depends on the values of constant $A$ and $B$. for a general values of $A$ and $B$ we can get a general complex solution like below assuming that $s^2+iBs+2A^2=(s-\omega_1)(s-\omega_2)$ in which $\omega_1,\omega_2=\frac{-i(B\pm \sqrt{B^2+8A^2})}{2}$
$$X(s)=\frac{\frac{-A^2}{(\omega_1-\omega_2)(\omega_1+iB)}}{s-\omega_1}+\frac{\frac{-A^2}{(\omega_2-\omega_1)(\omega_2+iB)}}{s-\omega_2}+\frac{\frac{1}{2}}{s+iB}\\ 
Y(s)=\frac{\frac{iA}{\omega_1-\omega_2}}{s-\omega_1}+\frac{\frac{iA}{\omega_2-\omega_1}}{s-\omega_2}\\
Z(s)=\frac{\frac{-A^2}{(\omega_1-\omega_2)(\omega_1+iB)}}{s-\omega_1}+\frac{\frac{-A^2}{(\omega_2-\omega_1)(\omega_2+iB)}}{s-\omega_2}+\frac{\frac{-1}{2}}{s+iB}$$
now we must take Inverse Laplace Transform from above equations using these facts:
$$\mathscr L\{e^{at}u(t)\}=\frac{1}{s-a}\iff \mathscr L^{-1}\{\frac{1}{s-a}\}=e^{at}u(t)$$
where $u(t)$ is Unit Step Function A.K.A. Heaviside Step Function. so writing time functions is left to you for example 
$$y(t) = (\frac{iA}{\omega_1-\omega_2}e^{\omega_1t}+\frac{iA}{\omega_2-\omega_1}e^{\omega_2t})u(t)$$
also you can investigate the last condition $|x(t)|^2+ |y(t)|^2+ |z(t)|^2=1$  holds in Laplace domain using this fact:
$$\mathscr L\{f(t)\}=F(s) \Rightarrow \mathscr L\{|f(t)|^2\}=\mathscr L\{f(t)f(t)^*\}=F(s)F(-s)$$
in which $^*$ stands for complex conjugate.
