I've been trying to solve the following equation $$ y(t)=-A\cdot\frac{\mathrm{d} y}{\mathrm{d} t}+B\cdot\left(\frac{\mathrm{d} y}{\mathrm{d} t}\right)^{2}+C \\ y(t=0)=y_{0}\\ $$ where $A$, $B$, and $C$ are positive numbers. The derivative at $t=0$ is also known but not really needed. I tried to solve it using scipy.integrate.solve_ivp but all the tutorials I watched don't have the derivate in the function and I can't figure out how to decouple it.

  • $\begingroup$ Are you sure that an ODE model is appropriate? The description looks more like a PDE is required. Also, height as function of velocity seems a little suspect, are you sure of the left side? $\endgroup$ Feb 23, 2023 at 13:20
  • $\begingroup$ I'm pretty sure its ok to describe the velocity as dh/dt the change in height over time (dh/dt) is the flowrate(Q) [m3/s] by the crosssection area(A) [m2] and you get dh/dt=Q/A=-v[m/s] As for the ode vs pde, the only derivative is water height by time to I think it is an ode Now I've noticed I'm missing the minus before the derivative, not sure it changes anything tho $\endgroup$ Feb 23, 2023 at 16:40
  • $\begingroup$ A,B,C just constants? Why not to solve the quadratic equation first to express y' = F(y), then go from there with any standard ODE method. $\endgroup$ Feb 23, 2023 at 17:46
  • $\begingroup$ @DavidKetcheson I only have first-order equations, its power of 2 not second derivative $\endgroup$ Feb 23, 2023 at 22:28
  • $\begingroup$ @MaximUmansky What do you mean? I have B*(y'(t))^2+A*y'(t)+(C-y(t)), for when y=0 i can solve like that, but i want to plot y(t), or am I missing something $\endgroup$ Feb 23, 2023 at 22:28

1 Answer 1


One preliminary comment is that supplying at the initial point both $y$ and $y'$ is redundant. If $y'(t_0)$ is given then obviously the ODE fully defines $y(t_0)$. On the other hand, if $y(t_0)$ is given then in general there are two different values of $y'(t_0)$ that satisfy the equation since the ODE can be interpreted as a quadratic equation for $y'$.

Extending this argument further, we solve for $y'$ in terms of $y$ and parameters $A$,$B$,$C$:

$$ y' = - \frac{A}{2B} \pm \frac{1}{2B} \sqrt{A^2 - 4 B (C-y)} $$

There are two branches here, and each of them defines an ODE in the canonical form $y_t=F(t,y)$ which can be solved with any standard method. Below is a short Python code solving the problem for illustrative values of the parameters: A=1, B=-1, C=1, y(0)=-100. There are two branches of the solution, and in this situation, one would usually need to invoke some physics arguments to establish if both branches are meaningful.

import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import solve_ivp

A=1; B=-1; C=1

#-two branches
Fp = lambda t, y: -0.5*A/B + (0.5/B)*np.sqrt(A*A - 4*B*(C-y))
Fm = lambda t, y: -0.5*A/B - (0.5/B)*np.sqrt(A*A - 4*B*(C-y))

t_eval = np.arange(0, 10., 0.1)
solp = solve_ivp(Fp, [0, 10.], [-100.], t_eval=t_eval)
solm = solve_ivp(Fm, [0, 10.], [-100.], t_eval=t_eval)

plt.plot(solp.t, solp.y[0])
plt.plot(solm.t, solm.y[0])

enter image description here

  • $\begingroup$ Thanks! I tried to supply as much info as possible, but should have added that A, B, and C should be positive which constricts my solution, at some points, it'll be ill-defined. the solution should only be the declining one. but much thanks! $\endgroup$ Feb 24, 2023 at 12:39
  • $\begingroup$ @BackSpace42 Glad to hear it was useful. If this answer solved your problem then please go ahead and accept it. $\endgroup$ Feb 24, 2023 at 16:56

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