Solving ODEs (solve_ivp) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of solve_ivp with simple ODE

What is the output array of y values when solving dy/dt = 2t with initial condition y(0)=1 over t in [0,1] using solve_ivp with default settings?

SciPy

from scipy.integrate import solve_ivp

def ode(t, y):
    return 2*t

sol = solve_ivp(ode, [0, 1], [1])
print(sol.y[0])

A

[1.         1.02040816 1.08163265 1.18367347 1.32653061 1.51020408
 1.73469388 2.      ]

B [1. 1. 1. 1. 1. 1. 1. 1.]

C [1. 2. 3. 4. 5. 6. 7. 8.]

D [1. 1.5 2.5 4. 6.5 10. 15.5 23.]

💡 Hint

Recall the analytical solution y = t^2 + 1 and that solve_ivp samples points automatically.

The ODE dy/dt = 2t integrates to y = t^2 + C. With y(0)=1, C=1. The solver returns approximate y values at chosen t points, close to t^2 + 1.

❓ data_output

intermediate

Number of time points in solve_ivp solution

Using solve_ivp to solve dy/dt = -3y with y(0)=5 over [0,1], how many time points does the solution contain by default?

SciPy

from scipy.integrate import solve_ivp

def ode(t, y):
    return -3*y

sol = solve_ivp(ode, [0, 1], [5])
print(len(sol.t))

A 10

B 100

C 50

D 5

💡 Hint

Check the default number of points solve_ivp evaluates if no t_eval is given.

By default, solve_ivp chooses about 10 time points adaptively unless t_eval is specified.

🔧 Debug

advanced

Identify the error in solve_ivp call

What error will this code raise? from scipy.integrate import solve_ivp def f(t, y): return t - y sol = solve_ivp(f, 0, [1])

SciPy

from scipy.integrate import solve_ivp

def f(t, y):
    return t - y

sol = solve_ivp(f, 0, [1])

A No error, runs successfully

B TypeError: 'int' object is not iterable

C ValueError: t_span must be a sequence of length 2

D TypeError: solve_ivp() missing 1 required positional argument: 'y0'

💡 Hint

Check the type and format of the second argument to solve_ivp.

The second argument t_span must be a tuple or list with two elements (start and end). Passing 0 (int) causes ValueError.

🚀 Application

advanced

Predict final value of logistic growth model

Using solve_ivp to solve the logistic growth ODE dy/dt = 0.5*y*(1 - y/10) with y(0)=1 over [0, 10], what is the approximate value of y at t=10?

SciPy

from scipy.integrate import solve_ivp

def logistic(t, y):
    return 0.5 * y * (1 - y / 10)

sol = solve_ivp(logistic, [0, 10], [1], t_eval=[10])
print(sol.y[0][0])

A 5.0

B 9.9

C 10.5

D 1.0

💡 Hint

The logistic model approaches the carrying capacity 10 over time.

The logistic growth approaches the carrying capacity 10, so y at t=10 is close to 10 but slightly less.

🧠 Conceptual

expert

Effect of changing 'method' parameter in solve_ivp

Which statement correctly describes the effect of changing the 'method' parameter in solve_ivp from 'RK45' to 'BDF'?

A Changing 'method' only affects the output format, not the solving process.

B 'BDF' always runs faster than 'RK45' regardless of the problem type.

C 'RK45' uses implicit methods suitable for stiff problems, 'BDF' uses explicit methods.

D 'BDF' is better for stiff ODEs and may use implicit methods, while 'RK45' is explicit and better for non-stiff problems.

💡 Hint

Consider the difference between explicit and implicit solvers and stiffness of ODEs.

'BDF' is an implicit solver designed for stiff ODEs, while 'RK45' is an explicit Runge-Kutta method suited for non-stiff problems.