1
$\begingroup$

I try to solve this equation:

enter image description here

And I got an answer:

DSolve[k'[t] == s k[t]^a - (n + b) k[t], k[t], t]

{{k[t] -> E^(-6 t) C[1]}}

But now I try this:

DSolve[{k'[t] == s k[t]^a - (n + b) k[t], k'[t] == 0}, k[t], t]

And got an error:

DSolve: There are fewer dependent variables than equations, so the system is overdetermined.

Please, help me. What should I do?

$\endgroup$
  • $\begingroup$ Please provide the values of s,a,n,b. $\endgroup$ – Ulrich Neumann Jan 21 '19 at 15:15
  • 3
    $\begingroup$ you meant k'[0] == 0 (not k'[t] == 0)? $\endgroup$ – kglr Jan 21 '19 at 15:16
  • $\begingroup$ s = 0.299, a = 0.35, n = 0.01, b = 0.1 $\endgroup$ – Тихон Сластен Jan 21 '19 at 15:18
3
$\begingroup$

Try

s = 0.299; a = 0.35; n = 0.01; b = 0.1; 
K = DSolveValue[{k'[t] == s k[t]^a - (n + b) k[t], k[0] == 0}, k ,t] // Chop
(*Function[{t},0.216305 2.71828^(-0.0715 t) (-1. + 1. 2.71828^(0.0715 t))(2.71828^(-0.0715 t) (-299. + 299. 2.71828^(0.0715 t)))^(7/13)]*)

Plot[K[t], {t, 0, 1}, AxesLabel -> {"t", "k[t]"}]

enter image description here

$\endgroup$

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.