How to solve this differential equation? (Bloch eq.)

Question

I have a set of equations: $$\frac{dM_{x}}{dt}=\omega M_{y}$$ $$\frac{dM_{y}}{dt}=\omega M_{x}$$ I solved the first one by substituting $M_{x}=M \cos(\omega t), M_{y}=M \sin(\omega t)\\$. I got the equation: $$\frac{dM}{M}=\omega \tan(\omega t),$$ which after integration is: $$M=M_{0} \cos(\omega t)\\$$ The solution of the second equation solved in a similar way is: $$M=M_{0} \sin(\omega t)\\$$

I know that the final solution should be $M=M_{0} \exp(i\omega t)\\$. My question is how to get such a solution? Everything would be perfect, if I had solutions for Mx and My and could just use the Eulers formula.

Answer 1

Add both equation $$\frac{dM_{x}}{dt}+\frac {dM_y}{dt}=\omega (M_{x}+M_{y})$$ $$\frac{d(M_{x}+M_y)}{dt}=\omega (M_{x}+M_{y})$$ Solve it then $$\int \frac{d(M_{x}+M_y)}{(M_{x}+M_{y})}= \omega \int dt $$ $$\ln (M_{x}+M_y)=\omega t+K$$ $$(M_{x}+M_y)=Ke^{\omega t}$$ take one equation and try to solve it to get $M_x$ or $M_y$ $$\frac{dM_{x}}{dt}=\omega M_{y}$$ $$\frac{dM_{x}}{dt}=\omega (Ke^{\omega t}-M_x)$$ $$\frac{dM_{x}}{dt}+\omega M_x=\omega Ke^{\omega t}$$ $$(M_{x}e^{\omega t})'=\omega Ke^{2\omega t}$$ $$M_{x}=\omega Ke^{-\omega t}\int e^{2\omega t} dt$$ $$M_{x}=K_1e^{-\omega t}+K_2e^{\omega t} $$ $$.................$$

Answer 2

$$\frac{d^2M_{x}}{dt^2}=\omega \frac{dM_{y}}{dt}=\omega^2 M_{x}$$

then

$$M_x=ae^{\omega t}+be^{-\omega t}.$$

There are no imaginary exponentials in this problem.