How can I write the differential equation below?

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seyyed Erfan ghoreyshipour
seyyed Erfan ghoreyshipour 2021 年 12 月 18 日
回答済み: seyyed Erfan ghoreyshipour 2021 年 12 月 19 日
I want to code the mathematical model below:
could you please help me out?
I want to estimate Q (flow) in this equation.
I also have omega=[7200,8800]
and can assign value to H.
After writing the code, I want to proceed with a simulink model.
Thank you for your help in advance

採用された回答

Torsten
Torsten 2021 年 12 月 18 日
編集済み: Torsten 2021 年 12 月 18 日
With
c1 = (b*omega + c*H)/L
c2 = (a + d*omega^2)/L
the solution Q(t) is
Q(t) = (-c1 + (c1+c2*Q0) * exp((t-t0)*c2)) / c2
where the initial condition is Q(t=t0) = Q0
  4 件のコメント
seyyed Erfan ghoreyshipour
seyyed Erfan ghoreyshipour 2021 年 12 月 19 日
Thank you!
It's great!
One question!
shouldn't it be c2 = ((a + d*omega^2)*Q)/L in the first equation?
Torsten
Torsten 2021 年 12 月 19 日
編集済み: Torsten 2021 年 12 月 19 日
No.
With
c1 = (b*omega + c*H)/L
c2 = (a + d*omega^2)/L
the differential equation reads
dQ/dt = c1 + c2*Q
thus
dQ/(c1+c2*Q) = dt
thus
integral_{Q0}^{Q} dQ/(c1+c2*Q) = integral_{t0}^{t} dt
thus
1/c2 * log((c1+c2*Q)/(c1+c2*Q0)) = t-t0
thus
exp(c2*(t-t0)) = (c1+c2*Q)/(c1+c2*Q0)
I leave it to you to solve for Q now.

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その他の回答 (1 件)

seyyed Erfan ghoreyshipour
seyyed Erfan ghoreyshipour 2021 年 12 月 19 日
Thank you master!

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