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Here is the code and the system, and it is supposedly analytically solvable.

$$\frac{dx(t)}{dt} = i(\frac{3w}{2}x + ke^{2iwt}y)$$ $$\frac{dy(t)}{dt} = i(kxe^{-2iwt} +\frac{3w}{2}y)$$

and my code attempt

DSolve[{x'[t] == I*((3*w/2) x[t] + y[t]*k*E^(2 I*w*t)), y'[t] == I*(k*x[t]*E^(-2*I*w*t) + 3*w*y[t]/2)}, {x[t], y[t]}, t]

I know it is something simple but I am not able to make it work. At least now it is not giving any errors, it simply does not compute anything. It just returns the same statement.

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  • $\begingroup$ In the second line, in the equation for $dy(t)/dt$, is there a $x$ missing in the first term on the right-hand side? If you edit your Mathematica code to reflect the $\LaTeX$ equations, there is a closed-form solution: DSolve[{x'[t] == I*((3*w/2) x[t] + y[t]*k*E^(2 I*w*t)), y'[t] == I*(k*E^(-2*I*w*t) + 3*w*y[t]/2)}, {x[t], y[t]}, t]. $\endgroup$
    – Roman
    Feb 12, 2019 at 21:22
  • $\begingroup$ Are these supposed to be a coupled equations? And do you have a solution already? Or are you trying to prove there is none? I'm suspicious that your x[t] should be a y[t].... $\endgroup$ Feb 13, 2019 at 10:14

1 Answer 1

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Yes, it appears you have an issue with your code, you have an extra x[t] in your second equation

DSolve[{x'[t] == I*((3*w/2) x[t] + y[t]*k*E^(2 I*w*t)), y'[t] == I*(k*E^(-2*I*w*t) + 3*w*y[t]/2)}, {x[t], y[t]},t] // FullSimplify

removing this and solving gives a solution:

$\left\{\left\{x(t)\to \frac{c_1 k e^{\frac{7 i t w}{2}}}{2 w}+c_2 e^{\frac{3 i t w}{2}}+\frac{4 k^2}{21 w^2},y(t)\to c_1 e^{\frac{3 i t w}{2}}-\frac{2 k e^{-2 i t w}}{7 w}\right\}\right\}$

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  • $\begingroup$ Seems that I messed up again. There's supposed to be an $x$ in there. I will add it to the LaTeX and edit the question $\endgroup$
    – The Bosco
    Feb 13, 2019 at 8:37

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