Showing a Function is a Solution

Show the code 
# This is a code cell that imports the necessary libraries for our session.
import numpy as np                        # NumPy for numerical computations
import scipy as sp                        # SciPy for scientific computing
import sympy as sym                       # SymPy for symbolic mathematics
import matplotlib as mpl                  # Matplotlib for plotting
import matplotlib.pyplot as plt           # Matplotlib pyplot interface
from scipy.integrate import solve_ivp     # ODE solver
mpl.rcParams['figure.dpi'] = 150
mpl.rcParams['axes.spines.top'] = False
mpl.rcParams['axes.spines.right'] = False

In this document, we will work through the details of showing that a particular function solves a particular differential equation. This is a common task in the study of differential equations, and it involves verifying that the function satisfies the equation when substituted into it.

Example 1

Here we will show both “by hand” and using SymPy that the function \(x(t) = \frac{1}{1+Ae^{-t}}\) is a solution to the ODE \(\frac{dx}{dt} = x(1-x)\).

By Hand

We want to verify that \[ x(t) = \frac{1}{1 + Ae^{-t}} \] satisfies \(\dfrac{dx}{dt} = x(1-x)\) for every value of the constant \(A\).

Step 1 — Compute the left-hand side \(\dfrac{dx}{dt}\).

Write \(x(t) = \bigl(1 + Ae^{-t}\bigr)^{-1}\) and apply the chain rule: \[ \frac{dx}{dt} = -\bigl(1 + Ae^{-t}\bigr)^{-2} \cdot \frac{d}{dt}\!\left(1 + Ae^{-t}\right) = -\frac{-Ae^{-t}}{\bigl(1+Ae^{-t}\bigr)^{2}} = \frac{Ae^{-t}}{\bigl(1+Ae^{-t}\bigr)^{2}}. \]

Step 2 — Compute the right-hand side \(x(1-x)\).

First observe that \[ 1 - x(t) = 1 - \frac{1}{1+Ae^{-t}} = \frac{1 + Ae^{-t} - 1}{1+Ae^{-t}} = \frac{Ae^{-t}}{1+Ae^{-t}}. \] Therefore \[ x(1-x) = \frac{1}{1+Ae^{-t}} \cdot \frac{Ae^{-t}}{1+Ae^{-t}} = \frac{Ae^{-t}}{\bigl(1+Ae^{-t}\bigr)^{2}}. \]

Step 3 — Compare.

\[ \frac{dx}{dt} = \frac{Ae^{-t}}{\bigl(1+Ae^{-t}\bigr)^{2}} = x(1-x). \checkmark \]

Since both sides are identical for every constant \(A\) and every \(t\), the function \(x(t) = \dfrac{1}{1+Ae^{-t}}\) is indeed a solution to \(\dfrac{dx}{dt} = x(1-x)\).

Note

Because \(A\) is an arbitrary constant, this is actually a one-parameter family of solutions — the general solution of this first-order ODE. A specific value of \(A\) is selected by imposing an initial condition.

Step 4 — Apply the initial condition \(x(0) = x_0\).

Setting \(t = 0\): \[ x(0) = \frac{1}{1 + A e^{0}} = \frac{1}{1+A} = x_0. \] Solving for \(A\): \[ 1 + A = \frac{1}{x_0} \quad \Longrightarrow \quad A = \frac{1}{x_0} - 1 = \frac{1-x_0}{x_0}. \]

Substituting back and simplifying: \[ x(t) = \frac{1}{1 + \dfrac{1-x_0}{x_0}\,e^{-t}} = \frac{x_0}{x_0 + (1-x_0)\,e^{-t}}. \]

The solution to the initial value problem \[ \frac{dx}{dt} = x(1-x), \qquad x(0) = x_0 \] is therefore \[ \boxed{x(t) = \frac{x_0}{x_0 + (1-x_0)\,e^{-t}}}. \]

Tip

Long-run behavior. Since \(e^{-t} \to 0\) as \(t \to \infty\), we have \(x(t) \to 1\) for any \(x_0 \in (0, 1)\). The constant solution \(x = 1\) is a stable equilibrium. Likewise \(x = 0\) is an unstable equilibrium: solutions starting slightly above zero still move away toward \(x = 1\).

Using SymPy

We now confirm the above calculation symbolically using SymPy.

Define the symbols and the candidate solution.

t, A, x0 = sym.symbols('t A x_0', real=True)

x = sym.S(1) / (1 + A * sym.exp(-t))

from IPython.display import Math, display
display(Math(r'x(t) = ' + sym.latex(x)))

\(\displaystyle x(t) = \frac{1}{A e^{- t} + 1}\)

Verify that \(\dfrac{dx}{dt} = x(1-x)\).

We compute both sides and subtract; if SymPy returns \(0\) the identity holds.

lhs = sym.diff(x, t)
rhs = x * (1 - x)

residual = sym.simplify(lhs - rhs)

print("LHS  (dx/dt)  =", sym.simplify(lhs))
print("RHS  (x(1-x)) =", sym.simplify(rhs))
print("LHS - RHS     =", residual)
LHS  (dx/dt)  = A*exp(t)/(A + exp(t))**2
RHS  (x(1-x)) = A*exp(t)/(A + exp(t))**2
LHS - RHS     = 0
Note

SymPy returns 0 for the residual, confirming algebraically that \(x(t) = \dfrac{1}{1+Ae^{-t}}\) satisfies the ODE.

Solve for \(A\) from the initial condition \(x(0) = x_0\).

ic_eq = sym.Eq(x.subs(t, 0), x0)
print("Initial condition equation:", ic_eq)

A_val = sym.solve(ic_eq, A)[0]
print("A =", A_val)
Initial condition equation: Eq(1/(A + 1), x_0)
A = (1 - x_0)/x_0

Write out the particular solution and simplify.

x_particular = sym.simplify(x.subs(A, A_val))
print("Particular solution: x(t) =", x_particular)

display(Math(r'x(t) = ' + sym.latex(x_particular)))
Particular solution: x(t) = x_0*exp(t)/(x_0*exp(t) - x_0 + 1)

\(\displaystyle x(t) = \frac{x_{0} e^{t}}{x_{0} e^{t} - x_{0} + 1}\)

Check once more that the particular solution satisfies the ODE.

lhs_p = sym.diff(x_particular, t)
rhs_p = x_particular * (1 - x_particular)
residual_p = sym.simplify(lhs_p - rhs_p)
print("Residual for particular solution:", residual_p)
Residual for particular solution: 0

Solution Curves for Various Initial Conditions

The plot below shows solution curves of the IVP \[ \frac{dx}{dt} = x(1-x), \qquad x(0) = x_0 \] for a range of initial values \(x_0\). The formula used is \(x(t) = \dfrac{x_0}{x_0 + (1-x_0)e^{-t}}\).

Show the code 
t_vals = np.linspace(-3, 6, 500)

# Initial conditions: below 0, near 0, spread through (0,1), above 1
x0_list = [-0.3, 0.02, 0.1, 0.3, 0.5, 0.7, 0.9, 0.98, 1.5, 2.5]

cmap = plt.cm.RdYlBu
colors = cmap(np.linspace(0.05, 0.95, len(x0_list)))

fig, ax = plt.subplots(figsize=(8, 5))

for x0_val, color in zip(x0_list, colors):
    # Avoid division by zero when x0 = 0 or x0 = 1
    denom = x0_val + (1 - x0_val) * np.exp(-t_vals)
    # Mask near-zero denominators for display stability
    with np.errstate(divide='ignore', invalid='ignore'):
        x_sol = np.where(np.abs(denom) > 1e-10, x0_val / denom, np.nan)
    ax.plot(t_vals, x_sol, color=color, lw=2,
            label=f'$x_0 = {x0_val}$')

# Equilibria
ax.axhline(1, color='black', linestyle='--', lw=1.2, label='Stable eq. $x=1$')
ax.axhline(0, color='gray',  linestyle='--', lw=1.2, label='Unstable eq. $x=0$')

ax.set_ylim(-0.6, 3.0)
ax.set_xlabel('$t$', fontsize=13)
ax.set_ylabel('$x(t)$', fontsize=13)
ax.set_title(r'Solutions of $\dfrac{dx}{dt} = x(1-x)$ for various $x_0$', fontsize=13)
ax.legend(fontsize=8, ncol=2, loc='upper left')
plt.tight_layout()
plt.show()
Figure 1: Solutions of \(dx/dt = x(1-x)\) for various initial conditions \(x_0\). All solutions with \(x_0 \in (0,1)\) increase toward the stable equilibrium \(x=1\) (upper dashed line), while solutions with \(x_0 > 1\) decrease toward it. The unstable equilibrium \(x=0\) (lower dashed line) repels nearby solutions.
Show the code 
import sys # sys for system-specific parameters and functions
print("Python version:", sys.version)
print('\n'.join(f'{m.__name__}=={m.__version__}' for m in globals().values() if getattr(m, '__version__', None)))
Python version: 3.14.4 | packaged by conda-forge | (main, Apr  8 2026, 02:33:53) [Clang 20.1.8 ]
numpy==2.4.3
scipy==1.17.1
sympy==1.14.0
matplotlib==3.10.8