Finding Particular Solutions via Fourier Series

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Recall

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: Variables that depend on one or more other variables (y).
• Independent variables: Variables upon which the dependent variables depend (x).
• Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

• The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

Second-order Linear Homogeneous ODE’s with Constant Coefficients

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0, where A and B are constants.

To solve this ODE, we seek two linearly independent solutions y₁(t) and y₂(t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c₁ and c₂ are two arbitrary constants determined by initial conditions. The key to solving the ODE is the characteristic equation, whose roots determine the behavior of the solutions.

Introduction to Fourier Series

In this section, we introduce the concept of Fourier series, a powerful mathematical tool used to represent periodic functions as sums of sine and cosine functions. This is particularly useful when dealing with inhomogeneous linear differential equations of the form: y’’ + ay’ + by = f(t) where

• f(t) is the unknown (input) function we aim to solve for, representing an external force or influence on the system.
• y(t) is the response or solution of the differential equation.
• The coefficients a and b describe characteristics of the system, such as damping or stiffness in physical systems.

Up to this point, we have often encountered input functions f(t) that are exponential, sine, or cosine functions. Although these functions might seem quite limited and special, they are in fact fundamental blocks because they allow us to build more complex periodic functions through superposition.

Due to the linear nature of the differential equation, we can decompose more complicated inputs into sums of sines and cosines, solve for their responses individually, and then combine the results to obtain the overall solution. This principle is known as superposition and is a cornerstone in solving linear differential equations.

A Fourier series expands any periodic function f(t) (with period 2π) as an infinite sum of sines and cosines. The general form of a Fourier series is f is periodic with period 2π, is $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where the Fourier coefficients $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

Understanding the convergence of Fourier series is essential when representing functions using trigonometric series. The Fourier series allows us to express periodic functions as infinite sums of sines and cosines, enabling us to analyze and solve differential equations with periodic inputs.

Finding Particular Solutions via Fourier Series

Let's explore the steps for finding a particular solution to the differential equation $x'' + w_o^2x = f(t)$ where the input f(t) is a square wave function. We will use Fourier series to express f(t) as an infinite sum of sine functions, which will allow us to find the particular solution x_p(t).

Defining the Square Wave Function f(t)

The function f(t) is defined on the interval [0, 2], and is periodic with period 2. Specifically, f(t) is given by: $f(t)= \begin{cases} 1, &0 < t < 1 \\ 0, &1 < t < 2 \end{cases}$. Refer to Figure i for a visual representation and aid in understanding it.

f(t) is periodic with period 2, meaning f(t +2k) = f(t) for all integers k.

f(t) is a square wave that alternates between 1 and 0 over intervals of length 1, repeating every 2 units. The graph consists of rectangles of height 1 and width 1.

To simplify the process of finding the Fourier series representation, we aim to work with an odd function. Odd functions have certain properties that make their Fourier series consist solely of sine terms, simplifying significantly our calculations.

We define a new function s(t) by shifting f(t) downward by $\frac{1}{2}: s(t) = f(t) -\frac{1}{2}$. Explicitly, s(t) is:

$s(t)= \begin{cases} 1/2, &0 < t < 1 \\ -1/2, &1 < t < 2 \end{cases}$

s(t) is also periodic with period 2: s(t + 2) = s(t). By subtracting 1/2 from the original function f(t), we make s(t) odd (s(-t) = -s(t)), which simplifies the process of constructing a Fourier series. It can be represented it using a Fourier sine series!

Consider the function g(u), defined on (0, π) and extended to an odd periodic function with period 2π: $g(u)= \begin{cases} 1, &0 < u < π \\ -1, &-π < u < 0 \end{cases}$

Fourier Sine Series of the Odd Extension. The Fourier series of g(u) is: $g(u) = \sum_{n=1, \text{n odd}}^\infty \frac{4}{nπ}sin(nu) = \frac{4}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nu)}{n}$

Now, to relate s(t) to g(u), observe that s(t) = $\frac{1}{2}g(u)$ and set u = π·t. This transformation maps the interval t ∈ [0 , 1] to u ∈ [0, π], s(t) = $\frac{1}{2}g(u) = \frac{1}{2}g(πt)$ and we cycle back to our original function f. Formally, we can express the original square wave function f(t) as:

f(t) = $\frac{1}{2}+ s(t) = \frac{1}{2}+\frac{1}{2}g(u) = \frac{1}{2}+\frac{1}{2}g(π·t) = \frac{1}{2}+\frac{1}{2}\frac{4}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n} = \frac{1}{2}+\frac{2}{π}\sum_{n=1, n~odd}^\infty \frac{sin(nπt)}{n}$. This is the Fourier series representation of the square wave function f(t).

Solving the Differential Equation

We return to our differential equation: $x’’+ w_o^2x = f(t)$. Our goal is to find a particular solution x_p(t).

We can calculate x_p if the right hand side of the previous equation is either cos(wt), sin(wt), or some multiple of them. The procedure is basically to convert the differential equation in a complex differential equation as we have previously done $x’’+ w_o^2x = cos(wt), x’’+ w_o^2x = sin(wt)$ and the solution is: $x_p = \frac{cos(wt)}{w_o^2-w^2}, \frac{sin(wt)}{w_o^2-w^2}$ respectively. Notice that if w=0, the formula still applies: $x_p = \frac{cos(0·t)}{w_o^2-0^2} = \frac{1}{w_o^2}$

Recall that if w₀ ≈ w, then the amplitude becomes very large and it produces the physical phenomenal called resonance.

Suppose $f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty (a_ncos(w_nt)+b_nsin(w_nt))$ where w_n = $\frac{nπ}{L}$ (period of f(t) is 2L).

Since the differential equation is linear and the operator x’’ + w₀²x is linear, we can use the principle of superposition. This means that the particular solution corresponding to the sum of inputs is the sum of the particular solutions corresponding to each input.

Finding Particular Solutions for Each Term

1. Constant term $\frac{a_0}{2}$ (w = 0). We solve x’’ + w₀²x = $\frac{a_0}{2}$. A particular solution is $x_p(t) = \frac{a_0}{2w_o^2}$
2. Sine terms: $sin(w_nt)$. We solve x’’ + w₀²x = $sin(w_nt)$. A particular solution is $x_p(t) = \frac{sin(wt)}{w_o^2-w^2}$
3. Cosine terms: $cos(w_nt)$. We solve x’’ + w₀²x = $cos(w_nt)$. A particular solution is $x_p(t) = \frac{cos(wt)}{w_o^2-w^2}$

$x_p = \frac{a_0}{2w_o^2}+\sum_{n=1}^\infty (\frac{a_ncos(w_nt)}{w_o^2-w_n^2}+\frac{b_nsin(w_nt)}{w_o^2-w_n^2})$

Particularly, if the input is the previous square wave function f(t), the differential equation is $x’’+ w_o^2x = f(t)$, and using the Fourier sine series expansion of f(t) (f(t) = $\frac{1}{2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n}$), we obtain the particular solution:

$x_p(t) =[\text{Observe that } w_n = nπ] \frac{1}{2w_o^2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n(w_o^2-(nπ)^2)}$

Analyze the Particular Solution

Let’s analyze the particular solution when w_o = 10.

Compute $w_o^2-(nπ)^2$:

1. n = 1: $w_o^2-(nπ)^2 = 100 -π^2 ≈100−9.87≈90.13$
2. n = 3: $w_o^2-(nπ)^2 = 100 -9π^2 ≈100−88.83≈11.17$
3. n = 5: $w_o^2-(nπ)^2 = 100 -25π^2 ≈100−246.74≈−146.74$

For n=5 and higher odd values of n, the denominator becomes negative and large in magnitude.

Using these values, we can approximate x_p(t) ≈ $\frac{1}{2·100} + \frac{2}{π}(\frac{sin(πt)}{1·90.13}+\frac{sin(3πt)}{3·11.17}+\frac{sin(5πt)}{5·(−146.74)}+···)$

Calculate the amplitudes:

1. n = 1, $\frac{2}{π}·\frac{1}{1·90.13}≈0.007$
2. n = 3, $\frac{2}{π}·\frac{1}{3·11.17}≈0.019$
3. n = 5, $\frac{2}{π}·\frac{1}{5·(−146.74)}≈−0.00087$

x_p(t) ≈ $0.005 + 0.007sin(πt) + 0.019sin(3πt) + −0.00087sin(7πt)+···$.

The terms with n = 1 and n = 3 have significantly larger amplitudes compared to higher n. Thus, the solution shows that the frequencies corresponding to n = 1 (πt) and n = 3 (3πt) are the most significant, as their amplitudes are much larger than the amplitudes of higher frequencies.

When w_o is close to one of these frequencies (w₀≈ π or w₀≈ 3π, even w₀≈ nπ), the denominator $w_o^2-(nπ)^2$ becomes small, causing the amplitude of the corresponding sine term to become large. This is the resonance phenomenon, where the system’s natural frequency w₀ matches one of the input frequencies, resulting in a large response.

The frequencies in f(t) are hidden, but the system picks out, favors, resonates or responds more strongly to the frequencies in the input that are close to its natural frequency. In our example with w₀ = 10, the frequency 3π ≈ 9.42 is close to w₀, resulting in a larger amplitude for n = 3: 0.019.

Alternative Method: Direct Substitution

Another way of solving this problem is to assume a particular solution of the form (the solution is similar to the input function): $x_p = c_0 + \sum_{n=1}^\infty c_nsin(nπt)$.

Then, we substitute this assumed solution into the original differential equation and match terms.

Computes the derivatives:

1. First derivative: $x_p’ = \sum_{n=1}^\infty c_n·nπ·cos(nπt)$
2. Second derivative: $x_p’’ = \sum_{n=1}^\infty -c_n(nπ)^2sin(nπt)$

Substituting into the differential equation $x’’+ w_o^2x = f(t)$, we have: $\sum_{n=1}^\infty -c_n(nπ)^2sin(nπt) + w_o^2(c_0 + \sum_{n=1}^\infty c_nsin(nπt)) = \frac{1}{2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n} ↭ w_o^2c_0 + \sum_{n=1}^\infty c_n(w_o^2-(nπ)^2)sin(nπt) = \frac{1}{2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n}$.

By matching terms, we find: $w_o^2c_0 = \frac{1}{2}↭[\text{Solving for }c_0]c_0 = \frac{1}{2w_o^2}$, and $c_n(w_o^2-(nπ)^2) = \frac{2}{π}·\frac{1}{n} ↭[\text{Simplifying and solving for } c_n] c_n = \frac{2}{πn}\frac{1}{w_o^2-(nπ)^2}$ (for odd n). This is the same solution we found earlier, $x_p(t) = c_0 + \sum_{n=1}^\infty c_nsin(nπt) = \frac{1}{2w_o^2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n(w_o^2-(nπ)^2)}$

Damped harmonic oscillator

The damped harmonic oscillator is a fundamental concept in physics and engineering, describing systems where both a restoring force (a force that acts to bring a system back to its equilibrium position after it has been displaced from that position, Hooke’s Law F_restoring = -kx) and a damping force (it is proportional to the velocity and opposes the motion, F_damping = -bx’(t)) act on an object. Such systems are prevalent in mechanical and electrical engineering, modeling phenomena like oscillating springs, circuits, and even financial markets.

The general equation of motion for a damped harmonic oscillator is: $mx’’ +bx’ + kx = F_{ext}(t)$ where

• m is the mass of the object.
• b is the damping coefficient (representing resistance such as friction or air resistance).
• k is the spring constant (representing the stiffness of the spring).
• F_ext is the external force acting on the system.
• x(t) is the displacement of the mass from its equilibrium position at time t.
• x′(t) and x′′(t) are the first and second derivatives of x(t) with respect to time, representing velocity and acceleration, respectively.

When there is no external force (i.e., F_ext = 0), the equation simplifies to the homogeneous case: $mx’’ +bx’ + kx = 0$. This is a second-order linear homogeneous differential equation that describes the motion of a damped oscillator without any external forces.

Solving the Differential Equation

To solve this differential equation, we begin by assuming a solution of the form x(t) = e^st, where s is a complex number to be determined. Substituting x(t) = e^st into the homogeneous equation gives: $m(s^2e^{st}) + b(se^{st})+ke^{st} = 0$.

Dividing both sides by e^st (which is never zero) simplifies the equation to the characteristic equation: p(s) = ms² + bs + k.

This is a quadratic equation in s, and its roots determine the behavior of the system.

Characteristic Equation and Roots

The characteristic equation can be written as: $s^2+\frac{b}{m}s + \frac{k}{m} = 0 ↭[\text{Complete the square}] (s+\frac{b}{2m})^2 -(\frac{b}{2m})^2 = -\frac{k}{m} ↭ (s+\frac{b}{2m})^2 = (\frac{b}{2m})^2 -\frac{k}{m}$.

Solving for s = $-\frac{b}{2m}±\sqrt{(\frac{b}{2m})^2 -\frac{k}{m}} = -\frac{b}{2m}±i\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}$

At this point, the behavior of the system depends on the sign of $(\frac{k}{m}-(\frac{b}{2m})^2)$.

Damping Regimes

• Underdamped systems ($\frac{k}{m}>(\frac{b}{2m})^2$). In this case, the system oscillates with a gradually decreasing amplitude due to damping because the roots of the characteristic equation are complex. The roots are: $s = \frac{-b}{2m}±iw_d$ where $w_d = \sqrt{\frac{k}{m}-(\frac{b}{2m})^2}$. This w_d is the damped frequency of oscillation.
• Overdamped system ($\frac{k}{m}<(\frac{b}{2m})^2$). In this case, the roots of the characteristic equation are real and negative (s = $-\frac{b}{2m}±\sqrt{(\frac{b}{2m})^2-\frac{k}{m}} = -\frac{b}{2m}±\sqrt{\frac{b^2-4mk}{4m^2}} = \frac{-b±\sqrt{b^2-4mk}}{2m}$), meaning the system does not oscillate. Instead, it slowly returns to equilibrium without oscillating, more slowly than in the critically damped case.
• Critically damped system ($\frac{k}{m}=(\frac{b}{2m})^2$). One real repeated root: $s = \frac{-b}{2m}$. This case represents the boundary between oscillatory and non-oscillatory motion. The system returns to equilibrium as quickly as possible without oscillating.

Solution for Each Damping Regime

Underdamped Case

Second-order Linear Homogeneous ODE’s with Constant Coefficients, Complex roots (Underdamped System). For the underdamped case, where the roots are complex $s = \frac{-b}{2m}±iw_d, w_d = \sqrt{\frac{k}{m}-(\frac{b}{2m})^2}$, the general solution to the homogeneous equation is:

$x_h = Ae^{\frac{-bt}{2m}}cos(w_dt-Φ)$ where

• A is the amplitude of the oscillation.
• Φ is a phase shift determined by the initial conditions.
• The exponential term $Ae^{\frac{-bt}{2m}}$ represents the damping (resistance), causing the amplitude of the oscillations to decay over time.
• The oscillations occur at the damped frequency ω_d.

We introduce two key quantities:

• Undamped Natural Frequency $w_n = \sqrt{\frac{k}{m}}$. This is the undamped natural frequency of the system. It's the frequency at which the system would oscillate if there were no damping (b = 0 ⇒ w_d = w_n).
• Damping ratio $ζ = \frac{b}{2w_nm} = \frac{b}{2\sqrt{km}}$. It measures the amount of damping in the system (how fast the amplitude decays) relative to critical damping.
• Damped Natural Frequency: $w_d = \sqrt{\frac{k}{m}-(\frac{b}{2m})^2} = \sqrt{w_n^2 -ζ^2w_n^2} = w_n\sqrt{1-ζ^2}$

Observe that $(\frac{b}{2m})^2 = \frac{ζ^2·4km}{4m^2}= ζ^2·\frac{k}{m} = ζ^2·w_n^2$

Another way of getting the same result is:

$w_d = \sqrt{\frac{k}{m}-(\frac{b}{2m})^2} = \sqrt{w_n^2-w_n^2(\frac{b}{2w_nm})^2} = w_n\sqrt{1-(\frac{b}{2w_nm})^2} = w_n\sqrt{1-ζ^2}$.

Using this notation, our homogeneous equation $s^2+\frac{b}{m}s + \frac{k}{m} = 0$ becomes $x’’ + 2ζw_nx’ + w_n^2x = 0$, and we can rewrite the solution for the underdamped case: $x_h = Ae^{\frac{-bt}{2m}}cos(w_dt-Φ) = Ae^{-w_nζt}cos(w_dt-Φ)$.

$+\frac{b}{m} = \frac{ζ2\sqrt{km}}{m} = ζ2\sqrt{\frac{km}{m^2}} = 2ζ\sqrt{\frac{k}{m}} = 2ζw_n$

$x_h = Ae^{-w_nζt}cos(w_dt-Φ)$ where $w_d = w_n\sqrt{1-ζ^2}$. This is the general form of the solution for an underdamped harmonic oscillator.

• A is the initial amplitude.
• Φ is the phase angle determined by initial conditions.
• The cos(w_dt-Φ) term represents oscillations at the damped natural frequency w_d.
• The $e^{-w_nζt}$ term represents an exponential decay of the amplitude over time. This is due to the damping force acting on the system, which removes energy from the oscillation.

The term $w_d = w_n\sqrt{1-ζ^2}$ is real and positive. The roots of the characteristic equation are complex $s = \frac{-b}{2m}±iw_d =[\frac{b}{m} =ζ2\sqrt{\frac{km}{m^2}} = 2ζ\sqrt{\frac{k}{m}} = 2ζw_n] -w_nζ±iw_d$. The imaginary part (iw_d) leads to oscillations, while the real part (^-b⁄_2m = -w_nζ) introduces an exponential decay factor. The system oscillates (due to the cosine term), but the amplitude decays over time due to damping (there's an exponentially decaying factor).

Critically damped (ζ = 1)

ζ = 1 ⇒ $1 = \frac{b}{2w_n·m} ⇒ w_n = \frac{b}{2m}$

$w_d = 0, s = \frac{-b}{2m} = -w_n$. The characteristic equation of the differential equation has a double root.

Second-order Linear Homogeneous ODE’s with Constant Coefficients, Two equal roots (Critically damped system). In this case, both roots are real and equal, meaning there is no imaginary component, so there is no oscillation (the solution contains no sine or cosine terms, which are responsible for oscillations in the underdamped case). The solution to the equation has the form: $x(t)=(C_1+C_2t)e^{\frac{-b}{2m}}$ where C₁ and C₂ are constants determined by initial conditions.

It is a purely decaying function. The system does not oscillate and returns to equilibrium as quickly as possible without overshooting.

Overdamped (ζ > 1)

The characteristic equation of the differential equation mx’’ + bx’ + kx = 0 has two distinct real roots: $s_1 = \frac{-b+\sqrt{b^2-4mk}}{2m}, s_2 = \frac{-b-\sqrt{b^2-4mk}}{2m}$.

Second-order Linear Homogeneous ODE’s with Constant Coefficients, Real and Distinct Roots (Overdamped System). With two distinct real roots (s₁ and s₂), both negative, the general solution for an overdamped system is: x(t) = C₁e^s₁t + C₂e^s₂t.

Since the damping coefficient b is large, there is no imaginary part, and the solution consists of two exponentially decaying terms. Therefore, the system exhibits no oscillatory behavior but returns to equilibrium more slowly than in the critically damped case. The motion is purely a smooth decay back to equilibrium.

Summary of Damping Regimes

Physical Interpretation of Damping

In a physical system, damping represents resistive forces like friction or air resistance. In the underdamped case, these forces are not strong enough to immediately stop the oscillations, but they gradually remove energy from the system. As a result, the system oscillates, but the amplitude gets smaller with each cycle, leading to a “decaying” oscillatory motion.

If a door closer (a system) is overdamped, the door will close very slowly and sluggishly, taking longer than necessary to shut, without oscillations.

In a door closer, critical damping ensures the door shuts smoothly and quickly without slamming or bouncing back. Damping is just enough to prevent oscillations, allowing the system to return to equilibrium quickly.

Finding the General Solution to y’’ + 3y = 2x

First, we are going to find a particular solution.

Step 1: Find the Fourier Series Representation of 2x on [−1, 1]

Since 2x is an odd function (i.e., f(−x) = -2x = −f(x)), its Fourier series expansion will contain only sine terms. This is known as the Fourier sine series, 2x = $\sum_{n=1}^\infty b_nsin(\frac{nπx}{L})$ where the coefficients b_n are given by:

$b_n = \frac{2}{L}\int_{0}^{L} f(x)sin(\frac{nπx}{L})dx = [\text{In our case, L = 1}] 2\int_{0}^{1} 2xsin(nπx)dx = 4\int_{0}^{1} xsin(nπx)dx$

Integration by parts. Let u = x⇒ du = dx, dv = sin(nπx)dx ⇒ v = $\frac{-1}{nπ}cos(nπx)$. I = uv −∫vdu

$\int_{0}^{1} xsin(nπx)dx = \frac{-x}{nπ}cos(nπx) + \frac{1}{nπ}\int cos(nπx)dx = \frac{-x}{nπ}cos(nπx) + \frac{1}{(nπ)^2}sin(nπx)\bigg|_{0}^{1} = $

• At t = 1, cos(nπ·1) = cos(nπ) = (-1)ⁿ, sin(nπ·1) = sin(nπ) = 0.
• At t = 0, cos(nπ·0) = cos(0) = 1, sin(nπ·0) = sin(0) = 0.

$b_n = 4([\frac{-1}{nπ}cos(nπ) + \frac{1}{(nπ)^2}sin(nπ)]-[\frac{0}{nπ}cos(0) + \frac{1}{(nπ)^2}sin(0)]) = 4([\frac{-1}{nπ}(-1)^n +0]-[0+0]) = \frac{4·(-1)^{n+1}}{nπ}$

The Fourier series for 2x on the interval [-1, 1] is: 2x = $\sum_{n=1}^\infty b_nsin(\frac{nπx}{L}) = [\text{In our case, L = 1}] \sum_{n=1}^\infty b_nsin(nπx) = \sum_{n=1}^\infty \frac{(-1)^{n+1}·4}{nπ}sin(nπx)$.

Step 2: Assume a Solution y(x) in Terms of a Fourier Sine Series

We assume that the particular solution y(x) can be expressed as a Fourier sine series: y = $\sum_{n=1}^\infty b_nsin(nπx)$. We choose a sine series because the right-hand side 2x is represented by sine terms, and we want our solution to match this structure.

We set a_n = 0 because the cosine terms do not appear in the Fourier series expansion of 2x, which contains only sine terms due to the odd symmetry of 2x over [−1, 1].

y = $\sum_{n=1}^\infty b_nsin(nπx)$

Step 3: Compute y′′+ 3y and substitute into our ODE

First derivative: y’(x) = $\sum_{n=1}^\infty b_nnπcos(nπx)$. Second derivative: $\sum_{n=1}^\infty -n^2π^2b_nsin(nπx)$

Substituting into our ODE: $\sum_{n=1}^\infty \frac{(-1)^{n+1}·4}{nπ}sin(nπx) = y’’ + 3y = \sum_{n=1}^\infty -n^2π^2b_nsin(nπx) + 3\sum_{n=1}^\infty b_nsin(nπx) = \sum_{n=1}^\infty (3-n^2π^2)b_nsin(nπx) ↭ \sum_{n=1}^\infty \frac{(-1)^{n+1}·4}{nπ}sin(nπx) = \sum_{n=1}^\infty (3-n^2π^2)b_nsin(nπx)$

Therefore: $\frac{(-1)^{n+1}·4}{nπ} = (3-n^2π^2)b_n↭[\text{Solving for }b_n] b_n = \frac{(-1)^{n+1}·4}{nπ(3-n^2π^2)}$

Step 4. Write the particular solution y(x): $y = \sum_{n=1}^\infty \frac{(-1)^{n+1}·4}{nπ(3-n^2π^2)}·sin(nπx)$

• Convergence of the Series: The series converges for all x in [−1, 1] due to the n³ behavior in the denominator, ensuring the terms decrease sufficiently fast.
• Homogeneous Solution: The general solution to the differential equation would include the homogeneous solution y_h(x) satisfying y′′ +3y =0. y(x)= y_h(x) + y_p(x)

Step 5. Solve the Homogeneous Equation

The corresponding homogeneous equation is: y′′ +3y =0. Assume a solution of the form y = e^rt. Substituting into the homogeneous equation: $r^2e^{rt} + 3e^{rt} = 0 ⇒[\text{Divide by } e^{rt}] r^2 + 3 = 0.$

Solve the characteristic equation for r: r² = -3 ⇒ $r = ±i\sqrt{3}$. These are complex conjugate roots.

Second-order Linear Homogeneous ODE’s with Constant Coefficients, Complex roots (Underdamped System). For complex roots r = α ± iβ, the general solution is: $y_h(x) = e^{αx}(C_1cos(βx)+ C_2sin(βx))$. Since α = 0 and β = $\sqrt{3}$, the homogeneous solution simplifies to: $y_h(x) = C_1cos(\sqrt{3}x) + C_2sin(\sqrt{3}x)$ where C₁ and C₂ are constants determined by initial conditions.

Step 6. Write the general solution

$y(x) = y_h(x) + y_p(x) = C_1cos(\sqrt{3}x) + C_2sin(\sqrt{3}x) + \sum_{n=1}^\infty \frac{(-1)^{n+1}·4}{nπ(3-n^2π^2)}·sin(nπx)$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare [18.03 Differential Equations, Spring 2006], YouTube by MIT OpenCourseWare.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.