Ask Your Question

Revision history [back]

A differential equation would involve the derivative z'(t) which you don't have in your equation(s).

If you substitute the value of r into the first equation you would have a single non-linear equation relating z and t. You may be able to solve this equation using the solve command. You don't have any unknown parameters in the expression for r or the equation involving z so you might not be able to solve the equation and have the values at t=1 that you specify be true.

You might try something like:

r,z,t = var('r,z,t')
equation = ((2.867e28) / (sqrt (r^3 - (2.1e13)^3 )) ) == (( (2.121e8)*(sqrt(2*(z^2) -1))) / ( z - sqrt (z^2 - 1)))
equation = equation.subs(r == ((1.97774e13)*(sqrt(sqrt(t)))) + ((3.55214e12)*(sqrt(t))))
solve(equation, t)

When I tried this I got:

Is  (2*z^2-1)*(z-sqrt(z^2-1))  positive, negative, or zero?

meaning that Maxima (the solver underneath Sage) can't solve the equation without more information.