Wave equation - Quantum Mechanics for Chemistry

What you will learn

Wave Equation Overview: The wave equation, a partial differential equation (PDE), fully describes the spatial and temporal evolution of waves.
Linearity of the Wave Equation: The wave equation is linear, meaning that any linear combination of solutions is also a solution. This allows the most general solution to be expressed as a linear combination of all possible solutions.
Boundary and Initial Conditions: To solve the wave equation for a specific physical system, boundary conditions must be specified. These include the values the wave function takes at the physical boundaries (e.g., $x = 0$ and $x = L$ ). Initial conditions with respect to time (e.g., at $t = 0$ ) may also be required.
Solution Method for 1D Guitar String: For a 1D guitar string, the wave equation is solved by separating variables, solving the resulting ordinary differential equations, and then applying boundary conditions at the two ends of the string.
Periodic Solutions: The solution for a 1D guitar string yields an infinite number of periodic solutions, parameterized by an integer $n$ .

Classical Wave equation¶

The wave equation is an example of a second-order PDE (partial differential equation). This PDE governs the behavior of the displacement $u(x,t)$ in time and space.

\frac{\partial^2 u(x,t)}{\partial x^2 }= \frac{1}{v^2}\frac{\partial^2 u(x,t)}{\partial t^2}

(1)

applied photoelectric — Figure 1:The classical wave equation describes any complicated wave in space and time, given the initial conditions.

Applications of the Classical Wave Equation: To illustrate the applications of the classical wave equation, we will solve it for a 1D guitar string. This provides a comprehensive mathematical description of the string’s behavior.
Predicting Evolution: By specifying arbitrary initial conditions, the wave equation allows us to precisely predict the evolution of the string over time and space.

Solving Wave Equation: The big picture¶

1. Boundary Conditions: To solve the wave equation for a specific physical situation, such as a guitar string fixed at both ends, we need to specify two boundary conditions. Mathematically, these are:

u(0, t) = 0 \quad \text{and} \quad u(L, t) = 0

(2)

where $u(x, t)$ represents the displacement of the string at position $x$ and time $t$ .
2. Separation of Variables: A common method to solve such equations is the technique of separation of variables. This technique assumes that $x$ and $t$ vary independently of each other, allowing us to express $u(x, t)$ as a product of two functions, each depending on only one variable. This lets us decompose the PDE into ordinary differential equations (ODEs).

u(x, t) = X(x) \cdot T(t)

(3)

3. Principle of superposition: The wave equation is linear, which you can see from the $u$ terms appearing to first order on both sides. Linearity means that a linear combination of two solutions $u_1$ and $u_2$ is also a solution. If you have $n$ particular solutions, then you write the general solution as a linear combination of those $n$ terms.

u = c_1 u_1+c_2u_2

(4)

Step 1: Plug the Product of Univariate Functions into the Wave Equation¶

Start with the 1D wave equation that can describe 1D guitar string:

\frac{1}{v^2}\frac{\partial^2 u(x,t)}{\partial t^2} = \frac{\partial^2 u(x,t)}{\partial x^2}

(5)

Substitute $u(x, t) = X(x)T(t)$ into the wave equation:

\frac{1}{v^2}\frac{\partial^2 X(x)T(t)}{\partial t^2} = \frac{\partial^2 X(x)T(t)}{\partial x^2}

(6)

After rearranging, we get:

\frac{X(x)}{v^2}\frac{\partial^2 T(t)}{\partial t^2} = T(t)\frac{\partial^2 X(x)}{\partial x^2}

(7)

Thus, we obtain two separate ordinary differential equations (ODEs) from the original partial differential equation (PDE):

\frac{1}{T(t)v^2}\frac{\partial^2 T(t)}{\partial t^2} = \frac{1}{X(x)}\frac{\partial^2 X(x)}{\partial x^2} = K

(8)

where $K$ is a constant.

Step 2: Solving Each Ordinary Differential Equation¶

After decomposing the PDE into ODEs, we solve each ODE separately:

\frac{\partial^2 T(t)}{\partial t^2} - K T(t) v^2 = 0

(9)

\frac{\partial^2 X(x)}{\partial x^2} - K X(x) = 0

(10)

Solving simple linear and homogeneous ODEs

Going from ODE to algebraic equations

To solve linear (no powers of $y$ ) and homogeneous (no constants, right-hand side is zero) ODEs, we plug in an exponential function, e.g. $y(x) \rightarrow e^{kx}$ .
This turns the ODE into a simple algebraic equation in which the only variable is $k$ .
In our case of the second-order wave equation, we will be dealing with simple quadratic equations:

a y^{''} + by^{'} + cy = 0

(11)

ak^2 + bk + c = 0,

(12)

The solutions of which are given as:

k_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

(13)

The most general solution is then written as a linear combination. We need extra information to determine $c_1$ and $c_2$ (e.g. boundary or initial conditions).

y = c_1e^{k_1}+c_2 e^{k_2}

(14)

Example ODE

\frac{d^2 y}{dx^2} +4y = 0

(15)

For instance, if we plug $y=e^{kx}$ into the following equation and cancel the exponentials on both sides, we get an algebraic equation in terms of $k$ .

k^2 +4 = 0

(16)

We get two solutions (which could be complex, real, or repeated), which we then use to write the general solution in the form of a superposition.

k=\pm 2i

(17)

y=c_1e^{+2i} + c_2e^{-2i}

(18)

Solving the spatial part: when $K > 0$

If $K > 0$ , we specify $K = \beta^2$ , where $\beta$ is a constant. The general solution for $X(x)$ is represented as a linear combination of particular solutions (principle of superposition):
$X(x) = c_1 e^{\beta x} + c_2 e^{-\beta x}$
(19)
Applying the boundary conditions $X(0) = X(L) = 0$ leads to $c_1 = c_2 = 0$ . Since a non-trivial linear combination of two exponential functions cannot be zero everywhere, this results in the trivial solution:
$X(x) = 0$
(20)
This implies that the string does not move, so there is no music!

Solving the spatial part: when $K < 0$

If $K < 0$ , we specify $K = -\beta^2$ , where $\beta$ is a constant. The general solution for $X(x)$ is:
$X(x) = c_1 e^{i \beta x} + c_2 e^{-i \beta x}$
(21)
Using Euler’s formula, this can be rewritten as:
$X(x) = c_1 (\cos(\beta x) + i \sin(\beta x)) + c_2 (\cos(\beta x) - i \sin(\beta x))$
(22)
Simplifying further:
$X(x) = (c_1 + i c_2) \cos(\beta x) + (c_1 - i c_2) \sin(\beta x)$
(23)
Let $A = c_1 + i c_2$ and $B = c_1 - i c_2$ :
$X(x) = A \cos(\beta x) + B \sin(\beta x)$
(24)
Applying the boundary conditions $X(0) = 0$ and $X(L) = 0$ :
- At $x = 0$ : $A \cdot 1 + B \cdot 0 = 0$ , so $A = 0$ .
- At $x = L$ : $X(L) = B \sin(\beta L) = 0$ .
For a non-trivial solution, $\sin(\beta L) = 0$ , which implies:
$\beta L = n \pi \quad \text{where} \quad n = 1, 2, 3, \ldots$
(25)
Hence:
$\beta = \frac{n \pi}{L}$
(26)
$X(x) = B \sin \left(\frac{n \pi}{L} x \right)$
(27)

Solving temporal part

Having identified the $K<0$ condition through the solution of the spatial part, we can now solve for $T(t)$ :

\frac{\partial^2 T(t)}{\partial t^2} - K v^2 T(t) = 0 \quad \text{with} \quad K = \beta^2

(28)

Substitute $K = \beta^2$ :

\frac{\partial^2 T(t)}{\partial t^2} - \beta^2 v^2 T(t) = 0

(29)

The solution is:

T(t) = D_n \cos(\omega_n t) + E_n \sin(\omega_n t)

(30)

\omega_n = \beta v = \frac{n \pi v}{L}

(31)

Solution for spatial part¶

Because of the boundary conditions ( $X(x)=X(L)=0$ ) imposed at the ends of the guitar string, we found an infinite number of solutions indexed by an integer $n$ . These terms are called normal modes and are visualized below.

X(x) = B \sin \left(\frac{n \pi}{L} x \right)

(32)

import numpy as np
import matplotlib.pyplot as plt

# Parameters
L = 1.0
B = 1.0
ns = [1, 2, 3, 4]
x = np.linspace(0, L, 800)

# 2x2 subplot
fig, axes = plt.subplots(2, 2, figsize=(8, 6))

for ax, n in zip(axes.ravel(), ns):
    Xn = B * np.sin(n * np.pi * x / L)
    ax.plot(x, Xn, color="C0")
    ax.axhline(0, color="black", linewidth=0.8)
    ax.set_title(f"n = {n}")
    ax.set_xlabel("x")
    ax.set_ylabel("X(x)")
    ax.set_xlim(0, L)

fig.suptitle("Normal modes of a fixed–fixed string")
fig.tight_layout(rect=[0, 0, 1, 0.96])

Solution for temporal part¶

For the temporal part, we do not have boundary conditions! Time is allowed to march forward into infinity. We find the solution in the form of a linear combination of sine and cosine functions with $\omega_n = \beta v = \frac{n \pi v}{L}$ , where $n = 1, 2, 3, \ldots$ indexes the normal modes.

T(t) = D_n \cos(\omega_n t) + E_n \sin(\omega_n t) = A_n \cos (\omega_n t + \phi_n)

(33)

In the second line we use a trig identity to express the sum of cosine and sine in terms of a single cosine. We still have two constants to determine.
The constants $A_n$ and $\phi_n$ in the temporal part are determined by initial conditions, e.g. what the amplitude and phase of the wave should be at the starting time $t=0$ .

Step 3 Full solution: A linear combination of normal modes¶

After solving the ODEs for the spatial and temporal parts, we now combine them into a full solution.
The complete description of any vibrational motion of the guitar string is given by the sum of the normal modes, $X_n(x)$ .
While the terms involving time, $T_n(t)$ , depend on how and where the string is plucked, the normal modes $X_n(x)$ remain the same for a given string.

About nodes¶

Places where the guitar string stays at the resting position of 0 are called nodes. We notice that the number of nodal points increases with $n$ .

For $n = 1$ : There are 0 nodes (excluding the endpoints). This is called the fundamental frequency or first harmonic.
For $n = 2$ : There is 1 node. This is known as the first overtone or second harmonic.
For $n = 3$ : There are 2 nodes. This is called the second overtone or third harmonic.
Note that the general solution to wave equation is expressed as a linear combination of all normal modes

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

# Parameters (reduced frames for faster export)
L = 1.0
v = 1.0
B = 1.0
A = {1: 1.0, 2: 1.0, 3: 1.0}
phi = {1: 0.0, 2: 0.0, 3: 0.0}

x = np.linspace(0, L, 600)

def X(n, x):
    return np.sin(n * np.pi * x / L)

def T(n, t):
    omega_n = n * np.pi * v / L
    return np.cos(omega_n * t + phi[n])

def u_mode(n, x, t):
    return B * X(n, x) * T(n, t)

def u_sum(modes, x, t):
    s = np.zeros_like(x)
    for n in modes:
        s += A[n] * X(n, x) * T(n, t)
    return s

omega1 = np.pi * v / L
T1 = 2 * np.pi / omega1
tmax = 2 * T1
frames = 120  # fewer frames
ts = np.linspace(0, tmax, frames)

fig, axes = plt.subplots(2, 2, figsize=(8, 6))
axes = axes.ravel()
titles = ["n = 1", "n = 3", "n = 1 + 2", "n = 1 + 2 + 3"]
lines = []
for ax, title in zip(axes, titles):
    ax.set_xlim(0, L)
    ax.set_ylim(-1.6, 1.6)
    ax.set_xlabel("x")
    ax.set_ylabel("u(x,t)")
    ax.set_title(title)
    ax.axhline(0, linewidth=0.8)
    line, = ax.plot([], [])
    lines.append(line)

fig.suptitle("Standing waves on a fixed–fixed string")

def init():
    for line in lines:
        line.set_data([], [])
    return lines

def update(idx):
    t = ts[idx]
    lines[0].set_data(x, u_mode(1, x, t))
    lines[1].set_data(x, u_mode(3, x, t))
    lines[2].set_data(x, u_sum([1, 2], x, t))
    lines[3].set_data(x, u_sum([1, 2, 3], x, t))
    return lines

ani = FuncAnimation(fig, update, frames=frames, init_func=init, blit=True, interval=1000*(tmax/frames))

plt.close(fig)  # Prevents static display of the last frame
HTML(ani.to_jshtml())

2D Membrane Vibrations¶

The wave function of a 2D membrane with fixed edges depends on two independent spatial variables, $x$ and $y$ . By applying the method of separation of variables, we decompose the wave function into three ordinary differential equations.

u(x, y, t) = X(x) Y(y) T(t)

(36)

For the spatial components $X(x)$ and $Y(y)$ , there are boundary conditions that must be satisfied at the fixed edges: Boundary condition along the $X$ -axis: $X(0) = X(L) = 0$ and Boundary condition along the $Y$ -axis: $Y(0) = Y(L) = 0$
The 2D normal mode is the product of two 1D modes. Since the dimensions are independent, we get integers $n$ for the $x$ -direction and $m$ for the $y$ -direction:

u(x, y, t) = \sum_n \sum_m A_{nm} \cos(\omega_{nm}t + \phi_{nm}) \sin\left(\frac{n\pi x}{a}\right) \sin\left(\frac{m\pi y}{b}\right)

(37)

The angular frequency $\omega_{nm}$ depends on the geometry of the membrane (with dimensions $a$ and $b$ ) and the mode numbers $n$ and $m$ :

\omega_{nm} = v\pi \left(\frac{n^2}{a^2} + \frac{m^2}{b^2}\right)^{1/2}

(38)

Where:

$v$ is the wave speed,
$n$ and $m$ are the mode numbers,
$a$ and $b$ are the dimensions of the membrane.

membrane vibrations — Figure 4:Vibrations of a 2D membrane.

Full derivation of 2D rectangular membrane problem

To extend the solution of the 1D guitar string problem to 2D, you can analyze a 2D vibrating membrane, such as a rectangular or circular drumhead. Here is a step-by-step solution for a 2D rectangular membrane:

Solution for a 2D Rectangular Membrane

Consider a rectangular membrane with dimensions $L_x$ and $L_y$ , fixed along its edges. The wave equation for the membrane is:

\frac{\partial^2 u(x, y, t)}{\partial t^2} = c^2 \left( \frac{\partial^2 u(x, y, t)}{\partial x^2} + \frac{\partial^2 u(x, y, t)}{\partial y^2} \right)

(39)

where $c$ is the wave speed.

Separation of Variables

Assume the solution can be written as a product of functions, each depending on only one variable:

u(x, y, t) = X(x) Y(y) T(t)

(40)

Substitute this into the wave equation:

\frac{d^2}{dt^2}[X(x) Y(y) T(t)] = c^2 \left( \frac{d^2}{dx^2}[X(x) Y(y) T(t)] + \frac{d^2}{dy^2}[X(x) Y(y) T(t)] \right)

(41)

This simplifies to:

X(x) Y(y) \frac{d^2 T(t)}{dt^2} = c^2 \left( T(t) \frac{d^2 X(x)}{dx^2} Y(y) + T(t) X(x) \frac{d^2 Y(y)}{dy^2} \right)

(42)

Divide through by $X(x) Y(y) T(t)$ :

\frac{1}{c^2} \frac{1}{T(t)} \frac{d^2 T(t)}{dt^2} = \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} + \frac{1}{Y(y)} \frac{d^2 Y(y)}{dy^2}

(43)

Since the left side depends only on $t$ and the right side depends only on $x$ and $y$ , each side must equal a constant, which we denote as $-\lambda$ :

\frac{1}{c^2} \frac{d^2 T(t)}{dt^2} = -\lambda

(44)

\frac{d^2 X(x)}{dx^2} + \frac{d^2 Y(y)}{dy^2} = -\lambda X(x)

(45)

Solving for $T(t)$

The ODE for $T(t)$ is:

\frac{d^2 T(t)}{dt^2} + \lambda c^2 T(t) = 0

(46)

This has the general solution:

T(t) = A \cos(\omega t) + B \sin(\omega t)

(47)

where $(\omega = \sqrt{\lambda} c)$ .

Solving for $X$ and $Y$

To solve the spatial part, we need to separate the problem into two parts:

\frac{d^2 X(x)}{dx^2} + \frac{d^2 Y(y)}{dy^2} = -\lambda X(x)

(48)

Assume:

\frac{d^2 X(x)}{dx^2} = -\alpha^2 X(x)

(49)

\frac{d^2 Y(y)}{dy^2} = -(\lambda - \alpha^2) Y(y)

(50)

where $(\lambda = \alpha^2 + \beta^2)$ .

So, we have:

\frac{d^2 X(x)}{dx^2} + \alpha^2 X(x) = 0

(51)

\frac{d^2 Y(y)}{dy^2} + \beta^2 Y(y) = 0

(52)

Boundary Conditions:

For a rectangular membrane with boundaries $x = 0$ , $x = L_x$ , $y = 0$ , and $y = L_y$ , we apply:

X(0) = X(L_x) = 0

(53)

Y(0) = Y(L_y) = 0

(54)

The solutions for $X(x)$ and $Y(y)$ are:

X(x) = C \sin\left(\frac{n \pi x}{L_x}\right)

(55)

Y(y) = D \sin\left(\frac{m \pi y}{L_y}\right)

(56)

where $n$ and $m$ are positive integers, and:

\alpha = \frac{n \pi}{L_x}, \quad \beta = \frac{m \pi}{L_y}

(57)

So, the general solution for $u(x, y, t)$ is:

u(x, y, t) = \sum_{n=1}^\infty \sum_{m=1}^\infty \left[ A_{nm} \cos\left(\omega_{nm} t\right) + B_{nm} \sin\left(\omega_{nm} t\right) \right] \sin\left(\frac{n \pi x}{L_x}\right) \sin\left(\frac{m \pi y}{L_y}\right)

(58)

where:

\omega_{nm} = c \sqrt{\left(\frac{n \pi}{L_x}\right)^2 + \left(\frac{m \pi}{L_y}\right)^2}

(59)

This solution describes the vibration modes of a 2D rectangular membrane, with each mode characterized by different integer values of $n$ and $m$ .

import numpy as np  
import matplotlib.pyplot as plt
import ipywidgets
from ipywidgets import interact, interactive, Dropdown
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

import plotly.graph_objects as go
from plotly.subplots import make_subplots

def membrane2d_mode(n=1, m=1, t=0):
    """
    Calculates the 2D grid of points (X, Y) and the normal mode displacement (Z) of a vibrating 
    rectangular membrane at a specific time t.

    Parameters:
    n (int): Mode number along the x-direction. Default is 1.
    m (int): Mode number along the y-direction. Default is 1.
    t (float): Time at which to evaluate the normal mode. Default is 0.

    Returns:
    X, Y, Z (numpy arrays): Grid of points in the X-Y plane and the corresponding 
                            membrane displacement Z(X, Y, t).
    """
    
    # Constants
    Lx, Ly = 1.0, 1.0  # Dimensions of the rectangular region
    v = 0.1            # Wave speed 
    omega = v * np.pi / Lx * (n**2 + m**2)  # Angular frequency for the normal mode
    
    # Create a spatial grid
    Nx, Ny = 100, 100  # Number of grid points in each dimension
    x, y = np.linspace(0, Lx, Nx), np.linspace(0, Ly, Ny)
    X, Y = np.meshgrid(x, y)

    # Compute the spatial part of the normal mode
    Z = np.sin(m * np.pi * X / Lx) * np.sin(n * np.pi * Y / Ly) * np.cos(omega * t)
    
    return X, Y, Z

def viz_membrane2d_plotly(n=1, m=1, t=0): 
    """
    Visualizes the 2D normal modes of a vibrating membrane on a square geometry using Plotly.
    
    This function displays both a 2D contour plot and a 3D surface plot of the membrane displacement.

    Parameters:
    n (int): Mode number along the x-direction. Default is 1.
    m (int): Mode number along the y-direction. Default is 1.
    t (float): Time at which to evaluate the normal mode. Default is 0.
    """
    
    # Get the membrane displacement at time t
    X, Y, Z = membrane2d_mode(n, m, t)
    
    # Create a Plotly subplot with 2 views: 2D contour and 3D surface
    fig = make_subplots(
        rows=1, 
        cols=2, 
        subplot_titles=('2D Contour Plot', '3D Surface Plot'),
        specs=[[{"type": "contour"}, {"type": "surface"}]]
    )
    
    # Add 2D contour plot to the left side
    fig.add_trace(
        go.Contour(x=X.flatten(), y=Y.flatten(), z=Z.flatten(), colorscale='RdBu'),
        row=1, col=1
    )

    # Add 3D surface plot to the right side
    fig.add_trace(
        go.Surface(x=X, y=Y, z=Z, colorscale='RdBu'),
        row=1, col=2
    )

    # Update the layout for better visualization
    fig.update_layout(
        title_text="2D Contour and 3D Surface Plots of Membrane Vibrational Normal Modes",
        width=1000,
        height=500
    )
    
    # Show the figure
    fig.show()

interact(viz_membrane2d_plotly, n=(1,1), m=(2,3), t=(0,100))

<function __main__.viz_membrane2d_plotly(n=1, m=1, t=0)>

The sound of music.¶

Music produced by musical instruments is a combination of sound waves with frequencies corresponding to a superposition of the normal modes (called harmonics or overtones in music) of those instruments.
Learn more from this series.

Figure 5:The size of a musical instrument reflects the range of frequencies it is designed to produce: smaller instruments produce higher frequencies, larger instruments lower frequencies.

Example Problems¶

Problem 1: Simple Harmonic Oscillator¶

Solve the ODE:

\frac{d^2 y}{dt^2} + \omega^2 y = 0

(60)

where $\omega$ is a constant.

Solution:

The characteristic equation is:

r^2 + \omega^2 = 0

(61)

Solving for $r$ :

r = \pm i \omega

(62)

Thus, the general solution is:

y(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t)

(63)

where $C_1$ and $C_2$ are constants determined by initial conditions.

Problem 2: Damped Oscillator¶

Solve the ODE:

\frac{d^2 y}{dt^2} + 2 \beta \frac{dy}{dt} + \omega^2 y = 0

(64)

where $\beta$ and $\omega$ are constants. Consider cases of $\beta^2 < \omega^2$ , $\beta^2 > \omega^2$ and $\beta^2 = \omega^2$

Solution:

The characteristic equation is:

r^2 + 2 \beta r + \omega^2 = 0

(65)

Solving for $r$ using the quadratic formula:

r = \frac{-2 \beta \pm \sqrt{(2 \beta)^2 - 4 \omega^2}}{2}

(66)

r = -\beta \pm \sqrt{\beta^2 - \omega^2}

(67)

Depending on the discriminant $\beta^2 - \omega^2$ , we have three cases:

Underdamping ( $\beta^2 < \omega^2$ ):
$y(t) = e^{-\beta t} [C_1 \cos(\omega_d t) + C_2 \sin(\omega_d t)]$
(68)
where $\omega_d = \sqrt{\omega^2 - \beta^2}$ .
Critical Damping ( $\beta^2 = \omega^2$ ):
$y(t) = (C_1 + C_2 t) e^{-\beta t}$
(69)
Overdamping ( $\beta^2 > \omega^2$ ):
$y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$
(70)
where $r_1$ and $r_2$ are the distinct real roots found from the quadratic formula.

Problem 3: ODE with Constant Coefficients¶

Solve the ODE:

\frac{d^2 y}{dt^2} - 4 y = 0

(71)

Solution:

The characteristic equation for this ODE is:

r^2 - 4 = 0

(72)

Solving for $r$ :

r^2 = 4

(73)

r = \pm 2

(74)

Thus, the general solution is:

y(t) = C_1 e^{2t} + C_2 e^{-2t}

(75)

where $C_1$ and $C_2$ are constants determined by initial conditions.

Problem 4: Simple 2nd order ODE¶

Solve the ODE:

\frac{d^2 y}{dt^2} + 9 y = 0

(76)

Solution:

The characteristic equation for this ODE is:

r^2 + 9 = 0

(77)

Solving for $r$ :

r^2 = -9

(78)

r = \pm 3i

(79)

Thus, the general solution is:

y(t) = C_1 \cos(3t) + C_2 \sin(3t)

(80)

where $C_1$ and $C_2$ are constants determined by initial conditions.

Problem 5: When ODE gives repeated roots¶

Solve the ODE:

\frac{d^2 y}{dt^2} - 6 \frac{dy}{dt} + 9 y = 0

(81)

Solution:

The characteristic equation for this ODE is:

r^2 - 6r + 9 = 0

(82)

Factoring:

(r - 3)^2 = 0

(83)

Thus:

r = 3 \text{ (repeated root)}

(84)

The general solution for repeated roots is:

y(t) = (C_1 + C_2 t) e^{3t}

(85)

where $C_1$ and $C_2$ are constants determined by initial conditions.