Im trying to solve a differential equation that i did on paper with the solve function but it gives me a huge exact answer all i want is an approximation how do i do that. heres my code.p = 161/880 Vo = 0 g = 32.2 Vt = g/p yo = 1200 syms t eq = ((-1/

how do i do that. heres my code.

p = 161/880
Vo = 0
g = 32.2
Vt = g/p
yo = 1200
syms t
eq = ((-1/p)*(Vo-Vt)*exp(-p*t))+(Vt*t)+yo+(1/p)*(Vo-Vt)
t = solve(eq,t)

heres what it returns
t =

(880*lambertw(0, -(340584721819893789*exp(84266571152752611/340584721819893760))/340584721819893760))/161 - 523394851880451/387028092977152

i just need an answer in seconds that would be nice

Answer 1

[Star Strider]

Use the vpa or double function to get an understandable result:

p = 161/880
Vo = 0
g = 32.2
Vt = g/p
yo = 1200
syms t
eq = ((-1/p)*(Vo-Vt)*exp(-p*t))+(Vt*t)+yo+(1/p)*(Vo-Vt);
t = solve(eq,t)
t_vpa = vpa(t)
t_double = double(t)

t_vpa =
- 2.1428498167327182026809021208394 + 8.0507403552584882828906649509827i

t_double =
    -2.1428e+000 +  8.0507e+000i

[Walter Roberson]

There are other solutions, for other branch cuts of LambertW, but they are all complex-valued as well.

[Star Strider]

Thank you, Walter.

It would be interesting to see the original circuit and differential equation. Time should not be complex.

[Walter Roberson]

I don't think that is from a circuit. The line

g = 32.2

is giving the standard gravity of Earth expressed in feet per second squared. Vo is initial velocity, Vt is termination velocity. yo looks like initial height. So this looks to me like a question involving a body being acted upon by gravity, but not in a straight forward manner. Possibly the exp() is some kind of air resistance term.

[Cam]

Thanks a ton yeah star strider is right time shouldn't be complex something is wrong with how i integrated the ode originally any way it does model someone falling from 1200 feet with a certain air resistance. Ill have to work with it till i get a better answer i didn't realize that it was giving me a complex answer.

[Star Strider]

@Cam — My pleasure. Unless you are not supposed to use the dsolve function, using it could help you find the error in your solution.

@Walter — The problem with having an engineering background that is almost entirely in electrical engineering is that ‘V’ means voltage, and late at night I look no further!