0
$\begingroup$

Consider the following system of differential equations

sys = {y'[t] == g*z[t], z'[t] == g*y[t], y[0] == y0, z[0] == z0};

which we can solve

DSolve[sys, {y, z}, t]

Now, what if the system has complex conjugates involved?

sys2 = {y'[t] == g*Conjugate[z[t]], z'[t] == g*Conjugate[y[t]], y[0] == y0, z[0] == z0};

In this case

DSolve[sys2, {y, z}, t]

will fail.

So my question is: How can I solve the system sys2? Also more general remarks about solving systems, where complex conjugates are involved are welcome.

P.S: We can assume g real and positive here, if that helps finding the answer.

$\endgroup$
  • 5
    $\begingroup$ isn't the standard method to solve the real part of the ode and the complex part (separately) then add solutions? i.e. if the solution to the real part is solR and the solution to the complex part is solX then the solution is solR+ I * solX $\endgroup$ – Nasser Jul 14 '16 at 13:54
3
$\begingroup$

Try this:

 DSolve[{y1'[t] == g*z1[t], y2'[t] == -g*z2[t], z1'[t] == g*y1[t], 
  z2'[t] == -g*y2[t], y1[0] == y01, y2[0] == y02, z1[0] == z01, 
  z2[0] == z02}, {y1[t], y2[t], z1[t], z2[t]}, t]



(*   {{y1[t] -> 1/2 E^(-g t) (y01 + E^(2 g t) y01 - z01 + E^(2 g t) z01), 
      z1[t] -> 1/2 E^(-g t) (-y01 + E^(2 g t) y01 + z01 + E^(2 g t) z01), 
      y2[t] -> 1/2 E^(-g t) (y02 + E^(2 g t) y02 + z02 - E^(2 g t) z02), 
      z2[t] -> -(1/2)
          E^(-g t) (-y02 + E^(2 g t) y02 - z02 - E^(2 g t) z02)}}  *)

Have fun!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.