Solving 2nd Order ODE for k: 20dx²/dt² + dx/dt + kx = 0

[nunos]

Find k:

$$20\frac{d^2x}{dt^2} + \frac{dx}{dt} + kx = 0$$

when t = 0
- x = 1
- dx/dt = 1

My attempt at this in maxima:

(%i1) 20*'diff( 'diff(x, t), t) + 'diff(x, t) + k*x = 0;
                                 2
                                d x   dx
(%o1)                        20 --- + -- + k x = 0
                                  2   dt
                                dt
(%i2) ode2(%, x, t);
Is  80 k - 1  positive, negative, or zero?

positive;
                                 k    1                     k    1
                            sqrt(- - ---) t            sqrt(- - ---) t
            - t/40               5   400                    5   400
(%o2) x = %e       (%k1 sin(---------------) + %k2 cos(---------------))
                                   2                          2
(%i3) ic2(%, x=1, t=0, 'diff(x,t) = 1);
                             k    1
                        sqrt(- - ---) t
            - t/40           5   400
(%o3) x = %e       (cos(---------------)
                               2
             k    1                                 k    1
        sqrt(- - ---) t                        sqrt(- - ---) t
             5   400              k    1            5   400
 - (sin(---------------) (20 sqrt(- - ---) sin(---------------)
               2                  5   400             2
            k    1                              k    1
       sqrt(- - ---) t                     sqrt(- - ---) t
            5   400            t/40             5   400
 + cos(---------------) + 40 %e    ))/(sin(---------------)
              2                                   2
                             k    1
                        sqrt(- - ---) t
           k    1            5   400
 - 20 sqrt(- - ---) cos(---------------)))
           5   400             2
(%i4) solve(%, k);
           sqrt(80 k - 1) t              t/20  2
(%o4) [sin(----------------) = - (sqrt(%e     x
                  40
                  t/40     sqrt(80 k - 1) t         t/20
 + ((320 k - 4) %e     cos(----------------) + 80 %e    ) x
                                  40
                  2 sqrt(80 k - 1) t           t/20      t/40          t/40
 + (4 - 320 k) cos (----------------) + 1600 %e    ) + %e     x + 40 %e    )
                           40
                         sqrt(80 k - 1) t
/(2 sqrt(80 k - 1)), sin(----------------) = 
                                40
        t/20  2                  t/40     sqrt(80 k - 1) t         t/20
(sqrt(%e     x  + ((320 k - 4) %e     cos(----------------) + 80 %e    ) x
                                                 40
                  2 sqrt(80 k - 1) t           t/20      t/40          t/40
 + (4 - 320 k) cos (----------------) + 1600 %e    ) - %e     x - 40 %e    )
                           40
/(2 sqrt(80 k - 1))]

What am I doing wrong or what am I not doing? Help is very appreciated.

Thanks.

 

[pbandjay]

Is this a homework question? From my initial attempt, I am getting two equations with three unknowns.

20x'' + x' + kx = 0. Try x(t) = e^rt, then 20r²e^rt + re^rt + ke^rt = 0 and 20r² + r + k = 0.

$$r_{1,2}=\frac{-1 \pm \sqrt{1 - 4(20)k}}{40}$$

If the discriminant is nonzero then

$$x(t) = c_1e^{r_1t} + c_2e^{r_2t}$$

$$x'(t) = c_1r_1e^{r_1t} + c_2r_2e^{r_2t$$

x(0) = 1 implies 1 = c₁ + c₂
x'(0) = 1 implies 1 = c₁r₁ + c₂r₂

It seems to me that there needs to be another constraint in order to solve for c₁, c₂ and k. Perhaps I am overlooking something, but I will keep trying.