Now is the time for us to shine. The time when our dreams are within reach and possibilities vast. Now is the time for all of us to become the people we have always dreamed of being. This is your world. You’re here. You matter. The world is waiting, Haley James Scott.

Arithmetic is where numbers fly like pigeons in and out of your head, Carl Sandburg.

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

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, $\frac{dy}{dx}$

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 y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

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

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

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 $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$

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.

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).

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.

Additionally, 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.

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 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, we 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 for 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).

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 w_{0} ≈ 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’’ + w_{0}^{2}x 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.

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

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

Let’s analyze the particular solution when w_{o} = 10.

Compute $w_o^2-(nπ)^2$:

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

- n = 1, $\frac{2}{π}·\frac{1}{1·90.13}≈0.007$
- n = 3, $\frac{2}{π}·\frac{1}{3·11.17}≈0.019$
- 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 (w_{0}≈ π or w_{0}≈ 3π, even w_{0}≈ 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 w_{0} 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 w_{0} = 10, the frequency 3π ≈ 9.42 is close to w_{0}, resulting in a larger amplitude for n = 3: 0.019.

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:

- First derivative: $x_p’ = \sum_{n=1}^\infty c_n·nπ·cos(nπt)$
- 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, $x_p(t) = c_0 + \sum_{n=1}^\infty c_nsin(nπt) = \frac{1}{2w_o^2}+\frac{2}{π}\sum_{n=1, \text{n odd}}^\infty \frac{sin(nπt)}{n(w_o^2-(nπ)^2)}$

The damped harmonic oscillator is a fundamental concept in physics and engineering, describing systems where 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 = -kx) and a damping force act on an object. The general equation of motion for a damped harmonic oscillator is: $mx’’ +bx’ + kx = F_{ext}$ 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.

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.

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) = ms^{2} + bs + k. This is a quadratic equation in s, and solving it will give us the behavior of the system.

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

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 $(\frac{k}{m}-(\frac{b}{2m})^2)$.

- 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.

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

Using these definitions, we can rewrite the damped frequency w_{d} as:

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

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
_{d}t-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 $w_d = w_m\sqrt{1-ζ^2}$ 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}⁄_{m}) 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).

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 C_{1} and C_{2} 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.

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 (s_{1} and s_{2}), both negative, the general solution for an overdamped system is: x(t) = C_{1}e^{s1t} + C_{2}e^{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**

Damping Ratio (ζ) | Discriminant (D) | Roots of Characteristic Equation | System Behavior |
---|---|---|---|

ζ<1 (Underdamped) | D < 0 | Complex conjugate roots $s = −ω_nζ ± iω_d$ | Oscillatory with decaying amplitude |

ζ=1 (Critically Damped) | D = 0 | Real repeated root s = −ωₙ | Fastest return to equilibrium without oscillation |

ζ>1 (Overdamped) | D > 0 | Real distinct roots s₁, s₂< 0 | Slow return to equilibrium without oscillation |

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.**

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) = −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 = $\frac{-1}{nπ}cos(nπx)$

$\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 n
^{3}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: r^{2} = -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: $y_h(x) = C_1cos(\sqrt{3}) + C_2sin(\sqrt{3})$ where C_{1} and C_{2} are constants determined by initial conditions.

Step 6. **Write the general solution**

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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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