And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous

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.

To solve this ODE, we seek two linearly independent solutions y_{1}(t) and y_{2}(t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c_{1} and c_{2} are two arbitrary constants determined by initial conditions.

Consider a mass-spring system where a mass m hangs on a spring. The position (displacement) of the mass at any time t is described by x(t), measured along the x-axis.

x-axis is vertical (it is not so in the diagram, sorry). When the mass is in equilibrium (denoted by 0), the spring is stretched. The origin of the x-axis is chosen so that the mass’s equilibrium position corresponds to x = 0. This means the spring is exerting an upward force to balance the downward gravitational force on the mass.

The movement of the mass is governed by Newton’s second law, which can be written as: **ma = mx’’ = F _{total} = -kx** (Hooke’s law states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance, F

Damping forces are a special type of force that are used to slow down or stop a motion. A dashpot is a device for cushioning or damping a movement (as of a mechanical part) to avoid shock.

This gives the equation of motion: mx’’ + cx’ + kx = 0 ↭[Dividing through by m, we get the simplified form:] $x’’ + \frac{c}{m}x’+ \frac{k}{m}x = 0$.

This is a second-order linear homogeneous ODE with constant coefficients that models the oscillatory motion of the spring system. To solve this equation, we need to find two independent solutions.

To solve the second-order linear ODE: y’’ + Ay’ + By = 0, we **assume** the solution is of the form $y = e^{rt}$ where r is a constant to be determined.

Substituting or plugging $y = e^{rt}$ into the differential equation gives:

$r^2e^{rt} + Are^{rt} + Be^{rt} = 0$ ⇒[Cancelling the common factor $e^{rt}$ (which is never zero), we obtain the **characteristic equation**:] $r^2 + Ar + B = 0$.

The form of the general solution to the ODE depends on the nature of the roots of this quadratic equation.

If the characteristic equation has two distinct real roots, say r_{1} and r_{2}, then the general solution is: y = $c_1e^{r_1t} + c_2e^{r_2t}$.

In mechanical or electrical systems, this is akin to a system where the damping is so strong that the system returns to equilibrium without oscillating. The resistance to motion is so high that the mass slowly creeps back to its equilibrium position (more slowly than if there were no damping) without oscillating. The terms $e^{r_1t}$ and $e^{r_2t}$ represent two exponentially decaying components.

Consider the equation: y’’ + 4y’ + 3y = 0.

The characteristic equation for the differential equation is: r^{2} + 4y + 3 = 0 ↭[Factoring gives] (r + 3)(r + 1) = 0. Thus, the roots are r_{1} = -3 and r_{2} = -1.

The general solution for the differential equation is: y = $c_1e^{-3t} + c_2e^{-t}$

Suppose we are given the initial condition y(0) = 1, y’(0) = 0.

To determine c_{1} and c_{2}, we first need to calculate the derivate of y: $y’ =-3c_1e^{-3t}- c_2e^{-t} $

Substituting the initial condition t = 0, y(0) = 1: 1 = c_{1} + c_{2}.

Similarly, substituting y’(0) = 0, 0 = -3c_{1} -c_{2}. Adding both 1 = -2c_{1}, so c_{1} = ^{-1}⁄_{2}, c_{2} = -3c_{1} = ^{3}⁄_{2}

The final solution to the differential equation is: y = $\frac{-1}{2}e^{-3t} + \frac{3}{2}e^{-t}$ (Refer to Figure 2 for a visual representation and aid in understanding it)

If the characteristic equation has complex roots, say r = a ± bi, the general solution involves sine and cosine terms.

We get two complex solutions, one of them is $y=e^{(a+bi)t} = e^{at}e^{bti} = e^{at}(cos(bt)+isin(bt))$⇒ Its real part is $e^{at}cos(bt)$ and its imaginary part is $e^{at}sin(bt)$ ⇒[Theorem, the real and imaginary parts of a complex solution are themselves solutions] $e^{at}cos(bt)$ and $e^{at}sin(bt)$ are **two independent real solutions** ⇒ Thus, the general solution is: y = $e^{at}(c_1cos(bt)+c_2sin(bt)) =$[Using the trigonometry identity a·cos(θ)+b·sin(θ) = c·cos(θ - Φ)] $e^{at}(c·cos(bt- Φ))$ where:

- $c = \sqrt{c_1^2+c_2^2}$
- $Φ = arctan(\frac{c_2}{c_1})$.

This corresponds to an underdamped system, where the system oscillates with decreasing amplitude.

Theorem (the real and imaginary parts of a complex solution are themselves solutions) If y(t) = u(t) + iv(t) is a complex solution to a differential equation y'' + Ay' + By = 0 with real coefficients A and B, then both u(t) and v(t) satisfy the differential equation (they are both real solutions).

Proof.

By assumption u(t) + iv(t) is a solution to a differential equation y’’ + Ay’ + By = 0, hence (u + iv)’’ + A(u + iv)’ + B(u + iv) = 0.

Compute derivatives and group real and imaginary parts taking into consideration that A and B are real numbers:

(u’’ + Au’ + Bu) + (v’’ + Av’ + Bv)i = 0 ⇒ u’’ + Au’ + Bu = 0, v’’ + Av’ + Bv = 0 ⇒Thus, u(t) and v(t) are real solutions.

Consider the equation y’’ + 4y’ + 5y = 0.

The characteristic equation for the differential equation is: r^{2} + 4r + 5 = 0 ↭[Using the quadratic formula:] $r = \frac{-4±\sqrt{-4}}{2} = -2±i$. Thus, the roots are: r = -2 ± i.

The complex roots indicate that the general solution can be expressed in terms of exponential and trigonometric functions. One complex solution is: y = $e^{-(2+i)t} = e^{-2t}[cos(t)+i·sin(t)]$ ⇒[Theorem. Its real and imaginary parts are two independent real solutions] $e^{-2t}cos(t), \text{ and } e^{-2t}sin(t)$ are two independent real solutions, and the real solution is finally y = $e^{-2t}(c_1cos(t) + c_2sin(t))$.

Applying the initial conditions y(0) = 1, y’(0) = 0.

y(0) = 1 = $e^{0}(c_1cos(0) + c_2sin(0)) = c_1,$ so c_{1} = 1.

Next, we we differentiate y(t): y’(t): y’(t) = $e^{-2t}[-2(c_1cos(t)+c_2sin(t)) + (-c_1sin(t)+c_2cos(t))]$

y’(0) = 0 ⇒ 0 = -2c_{1} + c_{2} = -2 + c_{2} ⇒ c_{2} = 2.

Therefore, the solution is: $e^{-2t}(cos(t) + 2sin(t)) = \sqrt{5}e^{-2t}cos(t- Φ)$ where $c = \sqrt{c_1^2+c_2^2} = \sqrt{1^2+2^2} = \sqrt{5}$ and Φ is a phase angle given by $Φ = arctan(\frac{c_2}{c_1}) = tan^{-1}(\frac{2}{1})≈ 63.43° [\text{or (degrees × π) ÷ 180 = radians}] π/3$ radians (Refer to Figure 3 for a visual representation and aid in understanding it)

In this case, y = $e^{at}(c_1cos(bt)+c_2sin(bt)) = e^{at}(ccos(bt- Φ))$, the system oscillates while gradually returning to equilibrium. The oscillations are damped over time, meaning they decrease in amplitude (the oscillations die out at a rate governed by $e^{at}$), but the system does not return to equilibrium immediately.

In the study of second-order linear homogeneous ordinary differential equations (ODEs) with constant coefficients y′’ +Ay′ +By = 0, we encounter different cases based on the nature of the roots of the characteristic equation. One such case is when the characteristic equation has two equal roots, which leads to a critically damped system.

The characteristic equation for this ODE is derived from assuming a solution of the form y = e^{rt} where r is a constant to be determined. Substituting this assumed solution into the ODE gives the characteristic equation: r^{2} + Ar + B = 0.

When this quadratic equation has two equal roots, it means that the discriminant is zero, Δ = A^{2}-4B = 0. This implies that the two roots of the characteristic equation are both r = −a, where a = $\frac{A}{2}.$ So the characteristic equation is: r^{2} + 2ar + a^{2} = 0.

This quadratic can be factored as (r+a)^{2} = 0. Thus, the characteristic equation has a
**repeated root** r = -a.

In general, for a second-order linear ODE, we need two independent solutions to form the general solution. However, when the characteristic equation has a repeated root, we only have one solution of the form: y_{1}(t) = e^{-at}. The challenge is that we need another linearly independent solution. To find the second solution, we use a special method.

Theorem (Reduction of Order). If y_{1} is a known solution to the differential equation y’’ + py’ + qy = 0, then there exists another linearly independent solution y_{2}(t) of the form y_{2}(t) = y_{1}u(t), where u(t) is a function to be determined.

This is called the reduction of order method.

Proof:

We know that y_{1}(t) is one solution. We assume that the second solution y_{2}(t) takes the form: y_{2}(t) = y_{1}(t)·u(t) where u(t) is an unknown function that we need to find.

We need to calculate the first and second derivatives of y_{2}(t).

- y’
_{2}(t) =[Using the product rule] y_{1}’(t)u(t)+y_{1}(t)u’(t) - y’’
_{2}(t) =[Again, applying the product rule to y’_{2}(t)] y’’_{1}(t)u(t) + 2y’_{1}(t)u’(t) + y_{1}(t)u’’(t)

Now, substitute into the original ODE y’’ + p(t)y’ + q(t)y = 0:

y’’_{1}(t)u(t) + 2y’_{1}(t)u’(t) + y_{1}(t)u’’(t) + p(t)(y_{1}’(t)u(t)+y_{1}(t)u’(t)) + q(t)y_{1}(t)·u(t) = 0

Next, group terms involving u(t), u’(t) and u’’(t): (y’’_{1}(t) + p(t)y_{1}’(t)+q(t)y_{1}(t))·u(t) +(2y’_{1}(t) + p(t)y_{1}(t))·u’(t) + y_{1}(t)u’’(t) = 0 ⇒[Simplify, y_{1}’’(t) + p(t)y_{1}’(t) + q(t)y_{1}(t) = 0 because y_{1}(t) is a known solution of the ODE. Therefore, the term involving u(t) vanishes:] 0·u(t) +(2y’_{1}(t) + p(t)y_{1}(t))u’(t) + y_{1}(t)u’’(t) = 0 ⇒(2y’_{1}(t) + p(t)y_{1}(t))u’(t) + y_{1}(t)u’’(t) = 0

Assuming y_{1}(t)≠ 0, divide the entire equation by y_{1}(t):

$u’’(t) + \frac{2y_1’(t)+p(t)y_1(t)}{y_1(t)}u’(t) = 0↭[\text{Let’s simplify the coefficients of u’(t)}] u’’(t) + (2\frac{y_1’(t)}{y_1(t)}+p(t))u’(t) = 0$

The equation becomes **a first-order linear ODE in u’(t)**. Let’s introduce the substitution v(t) = u’(t) to reduce the order:

$v’(t) + (2\frac{y_1’(t)}{y_1(t)}+p(t))v(t) = 0$

This is a first-order linear ODE for v(t), and it can be solved using an integrating factor. The integrating factor is: μ(t) = $e^{\int (2\frac{y_1’(t)}{y_1(t)}+p(t))dt}$

We can split the integral into two parts: μ(t) = $e^{\int (2\frac{y_1’(t)}{y_1(t)})dt}e^{\int p(t)dt} = e^{2ln|y_1(t)|}e^{\int p(t)dt} = e^{ln|y_1(t)|^2}e^{\int p(t)dt} = y_1(t)^2e^{\int p(t)dt}$

Now, we multiply the original ODE by the integrating factor μ(t):

$y_1(t)^2e^{\int p(t)dt}v’(t)+y_1(t)^2e^{\int p(t)dt}(2\frac{y_1’(t)}{y_1(t)}+p(t))v(t) = 0$

Simplify the left-hand side into the derivative of the product μ(t)v(t): $\frac{d}{dt}(μ(t)v(t)) = 0$

To solve for v(t), integrate both sides with respect to t: μ(t)v(t) = C_{1} where C_{1} is a constant of integration.

Now, solve for v(t): $v(t)=\frac{C_1}{μ(t)} = \frac{C_1}{y_1(t)^2e^{\int p(t)dt}}$, so this is the solution for v(t). Once v(t) is found, recall that
v(t)=u′(t). Integrate v(t) to find u(t), u(t) = $\int v(t)dt + C_2$ where C_{2} is another constant of integration. Finally, the second solution to the ODE is: y_{2}(t) = y_{1}(t)·u(t).

The general solution to the original second-order ODE is the linear combination of the two independent solutions y_{1}(t) and y_{2}(t): y(t) = C_{1}y_{1}(t) + C_{2}y_{2}(t) where C_{1} and C_{2} are arbitrary constants.∎

In our particular case, we already know that y_{1}(t) = e^{-at} is a solution. So, we can assume that the second solution is of the form y_{2}(t) = e^{-at}·u(t).

We now plug y_{2}(t) into the original ODE. To do this, we need to compute the first and second derivatives of y_{2}(t):

y’_{2}(t) =[Using the product rule] -ae^{-at}u + e^{-at}u’, y_{2}(t)’’ = a^{2}e^{-at}u -2ae^{-at}u’ + e^{-at}u''

Now, substitute y_{2}, y_{2}’ and y_{2}’’ into the original equation y’’ + 2ay’ + a^{2}y = 0, a^{2}e^{-at}u -2ae^{-at}u’ + e^{-at}u’’ -2a^{2}e^{-at}u +2ae^{-at}u’ + a^{2}e^{-at}u = 0 [Simplifying the equation] e^{-at}u’’ + (-2ae^{-at}+2ae^{-at})u’ + (a^{2}e^{-at}-2a^{2}e^{-at}+ a^{2}e^{-at})u = 0⇒e^{-at}u’’⇒[Since e^{-at} ≠ 0, we are left with the differential equation] u’’(t) = 0

u’’(t) = 0 ⇒[If u′′(t) = 0, then u′(t) must be a constant because the derivative of any constant function is zero] u’(t) = c_{1} where c_{1} is a constant ⇒[Integrating again] u(t) = c_{1}t + c_{2} where c_{2} is another constant that comes from the integration process.

Thus, the solution to u’’(t) = 0 is a linear function: u(t) = c_{1}t + c_{2}.

Thus, the second solution y_{2}(t) is y_{2}(t) = e^{-at}·u = e^{-at}·(c_{1}t + c_{2}). Since c_{2}e^{-at} is just a multiple of the first solution y_{1}(t) = e^{-at}, we **discard it for the sake of linear independence (so y _{2} is not a merely a scalar multiple of y_{1})** Therefore, the second independent solution is y

The general solution to the critically damped ODE is the linear combination of the two independent solutions: y(t) = c_{1}y_{1}(t) + c_{2}y_{2}(t) = c_{1}e^{-at} + c_{2}te^{-at} where c_{1} and c_{2} are arbitrary constants determined by initial conditions.

In physical systems (e.g., a mass-spring-damper system), critical damping occurs when the damping is just sufficient to bring the system back to equilibrium as quickly as possible without oscillating. The system returns to equilibrium without oscillating, but more slowly than in the overdamped case. The solution describes this behaviour:

**Exponential Decay**: The term e^{-at}causes the system to decay exponentially to zero.**Linear Term**: The presence of te^{-at}term shows that there is no oscillation, but the decay may be slower if the initial velocity is large.The critically damped response typically shows a smooth return to equilibrium without crossing the equilibrium position (no overshoot).

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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