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And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous
Some people drink from the fountain of knowledge, others just gargle, Robert Anthony
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., $\frac{dy}{dx} = 3x +5y$):
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:
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:
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.
A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:
This equation is homogeneous because the right-hand side is zero; there no external forcing terms (like a function of t) on the right-hand side.
This type of differential equation arises frequently in physics and engineering, modeling systems such as mechanical vibrations and electrical circuits.
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.
Consider a mechanical system where a mass m is attached to a spring and a damper (dashpot). The mass moves along the vertical x-axis, and its displacement from the equilibrium position at time t is denoted by x(t). The origin of the x-axis is chosen so that the mass’s equilibrium position corresponds to x = 0.
System Components:
Newton’s Second Law Applied. The total force acting on the mass is the sum of the spring and damping forces. The movement of the mass is governed by Newton’s second law, which can be written as: ma = mx’’ = Ftotal = Fspring + Fdamping = -kx -cx’ where m is the mass and a is the acceleration (Refer to Figure 1 for a visual representation and aid in understanding it)
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 both sides 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. It models the oscillatory motion of the mass-spring-damper 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. The nature of the roots leads to three cases:
When the characteristic equation has two distinct real roots, say r1 and r2, then the general solution is: y = $c_1e^{r_1t} + c_2e^{r_2t}$.
In mechanical or electrical systems, this scenario corresponds to an overdamped system.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 motions.
Consider the equation: y’’ + 4y’ + 3y = 0.
Step 1: Write the Characteristic Equation. The characteristic equation for the differential equation is: r2 + 4y + 3 = 0
Step 2: Solve for r. r2 + 4y + 3 = 0↭[Factoring gives] (r + 3)(r + 1) = 0. Thus, the roots are r1 = -3 and r2 = -1.
Step 3: Write the General Solution. The general solution for the differential equation y’’ + 4y’ + 3y = 0 is: y = $c_1e^{-3t} + c_2e^{-t}$
Step 4: Apply Initial Conditions. Suppose we are given the initial condition y(0) = 1, y’(0) = 0.
To determine c1 and c2, 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 = c1 + c2.
Similarly, substituting y’(0) = 0, 0 = -3c1 -c2. Adding both 1 = -2c1, so c1 = -1⁄2, c2 = -3c1 = 3⁄2
Step 5: Write the Particular Solution. 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 exponential and trigonometric functions, which model oscillatory behavior in physical systems such as underdamped mechanical vibrations.
A second-order linear homogeneous ODE with constant coefficients has the general form: y′′+Ay′ +By=0. To solve this ODE, we assume a solution of the form: y(t) = ert. 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:] $r^2 + Ar + B = 0$.
When the discriminant D = A2-4B is negative, the characteristic equation has complex roots: r = a ± bi
We get two complex solutions, one of them is $y=e^{(a+bi)t} = e^{at}e^{bti} =[\text{Using Euler’s Formula}] e^{at}(cos(bt)+isin(bt))$⇒ Its real part is $e^{at}cos(bt)$ and its imaginary part is $e^{at}sin(bt)$ ⇒[Theorem. Both the real u(t) and imaginary parts v(t) of a complex solution y(t) = u(t) + iv(t) to the differential equation y’’ + Ay’ + By = 0 with real coefficients A and B, 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:
The general solution (with complex roots) to the differential equation is: y = $e^{at}(c_1cos(bt)+c_2sin(bt))$ where:
This corresponds to an underdamped system, where the system oscillates with decreasing amplitude due to damping. The exponential term eat (with a < 0) represents the damping effect, while the trigonometric functions model the oscillatory behavior.
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 to the differential equation).
Proof.
By assumption u(t) + iv(t) is a solution to a differential equation y’’ + Ay’ + By = 0, substituting into the differential equation: (u + iv)’’ + A(u + iv)’ + B(u + iv) = 0.
Compute derivatives, grouping real and imaginary parts, and 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.
Step 1: Formulate the Characteristic Equation. The characteristic equation for the differential equation is: r2 + 4r + 5 = 0.
Step 2. Solve for Roots. r2 + 4r + 5 = 0↭[Using the quadratic formula:] $r = \frac{-4±\sqrt{-4}}{2} = -2±i$. Thus, the roots are: r = -2 ± i.
Step 3: Write the General Solution. 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 = $e^{-2t}(c_1cos(t) + c_2sin(t))$.
Step 4. Apply the initial conditions. Given y(0) = 1, y’(0) = 0.
y(0) = 1 = $e^{0}(c_1cos(0) + c_2sin(0)) = c_1,$ so c1 = 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 = -2c1 + c2 = -2 + c2 ⇒ c2 = 2.
Step 5: Write the Particular Solution. 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 $Φ = arctan(\frac{c_2}{c_1}) = tan^{-1}(\frac{2}{1})≈ 63.43° [\text{or (degrees × π) ÷ 180 = radians}] 1.107$ 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 (cos(bt- Φ)) 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 $e^{at}$), e.g., $e^{-2t}$ (a < 0) but the system does not return to equilibrium immediately.
This behavior is characteristic of an underdamped system in physics, such as a mass-spring-damper system where damping is present but not strong enough to prevent oscillations.
In the study of second-order linear homogeneous ordinary differential equations (ODEs) with constant coefficients, of the from y′’ +Ay′ +By = 0, different behaviors arise depending 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.
To solve the differential equation, we assume a solution of the form: y(t) = ert where r is a constant to be determined. Substituting this assumed solution into the ODE, we obtain: $r^2e^{rt}+re^{rt}+Be^{rt} = 0 ↭[e^{rt}≠0]$ r2 + Ar + B = 0. The nature of the roots depends on the discriminant of the characteristic equation: Δ = A2 -4B.
When this quadratic equation has two equal roots, it means that the discriminant is zero, Δ = A2-4B = 0. This implies that the two roots of the characteristic equation are real and equal. More specifically, r1 = r2 = $\frac{-A±Δ}{2} = \frac{-A}{2} = -a$, where a = $\frac{A}{2}⇒[A²-4B=0] A = 2a, 4B = A²=4a² ⇒B = a².$
Thus, the characteristic equation is: r2 + 2ar + a2 = 0. This quadratic can be factored as (r+a)2 = 0. Hence, the characteristic equation has a repeated root r = -a.
In general, for second-order linear ODEs, we need two independent solutions to construct the general solution. However, when the characteristic equation has a repeated root, we only have one solution of the form: y1(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 Method). If y1 is a known solution to the ordinary differential equation y’’ + py’ + qy = 0, then there exists another linearly independent solution y2(t) of the form y2(t) = y1u(t), where u(t) is a function to be determined.
This is called the reduction of order method.
Proof:
We know that y1(t) is one solution. We assume that the second solution y2(t) takes the form: y2(t) = y1(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 y2(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) + y1(t)u’’(t) + p(t)(y1’(t)u(t)+y1(t)u’(t)) + q(t)y1(t)·u(t) = 0
Next, group terms involving u(t), u’(t) and u’’(t): (y’’1(t) + p(t)y1’(t)+q(t)y1(t))·u(t) +(2y’1(t) + p(t)y1(t))·u’(t) + y1(t)u’’(t) = 0 ⇒[Simplify, y1’’(t) + p(t)y1’(t) + q(t)y1(t) = 0 because y1(t) is a known solution of the ODE. Therefore, the term involving u(t) vanishes:] 0·u(t) +(2y’1(t) + p(t)y1(t))u’(t) + y1(t)u’’(t) = 0 ⇒(2y’1(t) + p(t)y1(t))u’(t) + y1(t)u’’(t) = 0
Assuming y1(t)≠ 0, divide the entire equation by y1(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) = C1 where C1 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 C2 is another constant of integration. Finally, the second solution to the ODE is: y2(t) = y1(t)·u(t).
The general solution to the original second-order ODE is the linear combination of the two independent solutions y1(t) and y2(t): y(t) = C1y1(t) + C2y2(t) where C1 and C2 are arbitrary constants.∎
In our particular case, we already know that y1(t) = e-at is a solution. So, we can assume that the second solution is of the form y2(t) = e-at·u(t).
We now plug y2(t) into the original ODE. To do this, we need to compute the first and second derivatives of y2(t):
y’2(t) =[Using the product rule] -ae-atu + e-atu’, y2(t)’’ = a2e-atu -2ae-atu’ + e-atu''
Now, substitute y2, y2’ and y2’’ into the original equation y’’ + 2ay’ + a2y = 0, a2e-atu -2ae-atu’ + e-atu’’ -2a2e-atu +2ae-atu’ + a2e-atu = 0 [Simplifying the equation] e-atu’’ + (-2ae-at+2ae-at)u’ + (a2e-at-2a2e-at+ a2e-at)u = 0⇒e-atu’’ = 0⇒[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) = c1 where c1 is a constant ⇒[Integrating again] u(t) = c1t + c2 where c2 is another constant that comes from the integration process.
Thus, the solution to u’’(t) = 0 is a linear function: u(t) = c1t + c2.
Thus, the second solution y2(t) is y2(t) = e-at·u = e-at·(c1t + c2). Since c2e-at is just a multiple of the first solution y1(t) = e-at, we discard it for the sake of linear independence (so y2 is not a merely a scalar multiple of y1) Therefore, the second independent solution is y2(t) = te-at (For simplicity and without loss of generality, we can set c1 = 1).
The general solution to the critically damped ODE is the linear combination of the two independent solutions: y(t) = c1y1(t) + c2y2(t) = c1e-at + c2te-at where c1 and c2 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:
The critically damped response typically shows a smooth return to equilibrium without crossing the equilibrium position (no overshoot). It is common in mechanical systems (e.g., mass-spring-damper systems) designed to avoid oscillations, such as car suspension systems, control systems, and electronics.
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