1 | initial version |

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.

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