Second-order Linear ODE's. Two equal roots (Critically damped system).

Some people drink from the fountain of knowledge, others just gargle, Robert Anthony

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, $\frac{dy}{dx}$

Basic linear ODE

In mathematics, a linear ordinary differential equation (ODE) is an equation involving an unknown function and its derivatives, characterized by the property that both the function and its derivatives appear linearly. This means that the unknown function and its derivatives are raised only to the first power. They are never multiplied together, nor are they composed within nonlinear functions (such as $\sin(y)$ or $e^y$). The general form for a first-order linear ODE is: a(x)y’ + b(x)y = c(x).

Second-order Linear Homogeneous ODE’s with Constant Coefficients

In mathematics, a second-order linear homogeneous differential equation ODE with constant coefficients is an equation of the form: ay'' + by' + cy = 0 where:

• y(t) 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, respectively.
• t is the independent variable,
• a,b, and c are constant real numbers, with $a \ne 0$.

This equation is characterized by:

• Second-Order: The highest derivative present is the second derivative.
• Linear: y and its derivatives are all of the first power, and there are no products of y with its derivatives.
• Homogeneous: The right-hand side is zero, meaning there are no external forcing functions acting on the system.

This type of differential equation arises frequently in physics and engineering, modeling systems such as mechanical vibrations (mass-spring-damper) and electrical circuits (LRC).

The Principle of General Solution

To solve this ODE, we must find two linearly independent solutions $y_1(t)$ and $y_2(t)$. The fundamental theorem of linear ODEs states that the general solution is a linear combination of these two: $c_1y_1 + c_2y_2$ where c₁ and c₂ are two arbitrary constants determined by initial conditions of the system. Because a second-order equation involves two derivatives, we mathematically require two independent pieces of information (and thus two independent solutions) to describe the full spectrum of possible behaviors.

Two equal roots (Critically damped system)

In the study of second-order linear homogeneous ordinary differential equations (ODEs) with constant coefficients, taking 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 models a critically damped system.

To solve this differential equation, we assume a solution of the form: y(t) = e^rt where r is a constant to be determined. Substituting this assumed solution into the ODE, we obtain: $r^2e^{rt}+re^{rt}+Be^{rt} = 0$.

Sinc3 $e^{rt} \ne 0$, we can divide the entire equation by $e^{rt}$ to yield the characteristic equation: $r^2 + Ar + B = 0$. The nature of the roots depends on the discriminant of the characteristic equation: $\Delta = A^2 -4B$.

When this quadratic equation has two equal real roots, it means that the discriminant must be zero, $\Delta = A^2 -4B = 0$. This implies that the two roots of the characteristic equation are real, identical, and can be expressed as: $r_1 = r_2 = \frac{-A \pm \Delta}{2} = \frac{-A}{2} = -a$.

a = $\frac{A}{2}$. Because A²-4B = 0, 4B = A² = 4a² so it follows that A = 2a and B = a².

Thus, the characteristic equation is: r² + 2ar + a² = 0. This elegantly factors into (r+a)² = 0, mathematically confirming that the characteristic equation has a repeated root at r = −a.

The Challenge of Finding Two Independent Solutions

For second-order linear ODEs, we absolutely need two linearly independent solutions to construct the general solution. However, because our characteristic equation only yielded a single repeated root, we currently only have one solution: $y_1(t) = e^{-at}$.

The challenge is that we need another linearly independent solution. To find the second solution, we use a special method.

Theorem (Reduction of Order Method). If $y_1(t)$ is a known, non-trivial solution to the ordinary differential equation $y'' + p(t)y' + q(t)y = 0$, then there exists another linearly independent solution $y_2(t)$ of the form $y_2(t) = y_1(t)u(t)$, where $u(t)$ is a function to be determined.

Proof:

Step 1. Define the assumed solution and its derivatives

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).

• First derivative: $y'_2(t) = y'_1(t)u(t) + y_1(t)u'(t)$
• Second derivative: Applying the product rule again and grouping terms yields: $y''_2(t) = y''_1(t)u(t) + 2y'_1(t)u'(t) + y_1(t)u''(t)$

Step 2. Substitute into the original ODE y’’ + p(t)y’ + q(t)y = 0

Substitute $y_2$, $y'_2$, and $y''_2$ back into the general ODE $y'' + p(t)y' + q(t)y = 0$: $\big[y''_1(t)u(t) + 2y'_1(t)u'(t) + y_1(t)u''(t)\big] + p(t)\big[y'_1(t)u(t) + y_1(t)u'(t)\big] + q(t)\big[y_1(t)u(t)\big] = 0$

Step 3. Group by the unknown function $u(t)$

Next, we rearrange the terms to group $u(t)$, $u'(t)$, and $u''(t)$ together: $u(t)\big[y''_1(t) + p(t)y'_1(t) + q(t)y_1(t)\big] + u'(t)\big[2y'_1(t) + p(t)y_1(t)\big] + u''(t)\big[y_1(t)\big] = 0$

Because $y_1(t)$ is already a known solution to the original ODE, the entire bracket multiplied by $u(t)$ equals zero. The equation drastically simplifies to: $u''(t)y_1(t) + u'(t)\big[2y'_1(t) + p(t)y_1(t)\big] = 0$

Step 4. Reduce the order

Assuming $y_1(t) \neq 0$, we divide the entire equation by $y_1(t)$: $u''(t) + \left(2\frac{y'_1(t)}{y_1(t)} + p(t)\right)u'(t) = 0$.

Notice that $u(t)$ is gone. By introducing a substitution $v(t) = u'(t)$ (and therefore $v'(t) = u''(t)$), we successfully reduce the second-order ODE into a first-order linear ODE: $v'(t) + \left(2\frac{y'_1(t)}{y_1(t)} + p(t)\right)v(t) = 0$

Step 5. Solve using an integrating factor

This first-order ODE can be solved using an integrating factor, $\mu(t)$: $\mu(t) = e^{\int \left(2\frac{y'_1(t)}{y_1(t)} + p(t)\right) dt}$

We can split this integral into two manageable parts:

$$\mu(t) = e^{\int 2\frac{y'_1(t)}{y_1(t)} dt} \cdot e^{\int p(t) dt} = e^{2 \ln|y_1(t)|} \cdot e^{\int p(t) dt} = y_1(t)^2 e^{\int p(t) dt}$$

Multiplying our first-order ODE by $\mu(t)$ turns the left-hand side into the exact derivative of the product $\mu(t)v(t)$: $\frac{d}{dt}\big(\mu(t)v(t)\big) = 0$

Integrating both sides with respect to $t$ gives $\mu(t)v(t) = C_1$.

Solving for $v(t)$ yields: $v(t) = \frac{C_1}{\mu(t)} = \frac{C_1}{y_1(t)^2 e^{\int p(t) dt}}$

Since $v(t) = u'(t)$, we integrate one final time to find $u(t)$: $\boxed{u(t) = \int \frac{C_1}{y_1(t)^2 e^{\int p(t) dt}} dt + C_2}$ 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.∎

Applying the Result to the Critically Damped System

For our specific critically damped system, the characteristic equation yields a repeated root, giving us the differential equation $y'' + 2ay' + a^2y = 0$, meaning $p(t) = 2a$. Our known first solution is $y_1(t) = e^{-at}$. We can find the second linearly independent solution using two different approaches.

Method 1: Using the Reduction of Order Formula

Let’s plug our knowns into our Reduction of Order formula for $v(t)$

1. Calculate the integrating factor’s exponent: $e^{\int p(t) dt} = e^{\int 2a dt} = e^{2at}$
2. Square $y_1(t)$: $y_1(t)^2 = (e^{-at})^2 = e^{-2at}$.
3. Find $v(t)$: $v(t) = \frac{C_1}{e^{-2at} \cdot e^{2at}} = \frac{C_1}{1} = C_1$
   If $u'(t) = C_1$, then simply integrating gives us $u(t) = C_1t + C_2$.
4. To find the most fundamental second independent solution, we can set $C_1 = 1$ and $C_2 = 0$, giving us $u(t) = t$. Therefore, our second solution is: $y_2(t) = y_1(t)u(t) = t e^{-at}$
5. Conclusion: The general solution to the critically damped second-order ODE is the linear combination of the two independent solutions we have just proven: $\boxed{y(t) = c_1 e^{-at} + c_2 t e^{-at}}$

Method 2: Direct Substitution (From Scratch)

Alternatively, knowing that $y_1(t) = e^{-at}$ is a solution, we can assume the second solution takes the form $y_2(t) = e^{-at}u(t)$ and plug it directly into the original ODE.

Step 1. Compute the derivatives using the product rule:

• $y_2(t) = e^{-at}u$
• $y'_2(t) = -ae^{-at}u + e^{-at}u'$
• $y''_2(t) = a^2e^{-at}u - 2ae^{-at}u' + e^{-at}u''$

Step 2. Substitute into the original equation.

Substitute $y_2$, $y'_2$, and $y''_2$ into $y'' + 2ay' + a^2y = 0$: $(a^2e^{-at}u - 2ae^{-at}u' + e^{-at}u'') + 2a(-ae^{-at}u + e^{-at}u') + a^2(e^{-at}u) = 0$

Step 3. Expand and group terms: $(a^2e^{-at}u - 2ae^{-at}u' + e^{-at}u'') - 2a^2e^{-at}u + 2ae^{-at}u' + a^2e^{-at}u = 0$

Now, group the terms by $u''$, $u'$, and $u$: $e^{-at}u'' + (-2ae^{-at} + 2ae^{-at})u' + (a^2e^{-at} - 2a^2e^{-at} + a^2e^{-at})u = 0$

Notice how perfectly the terms cancel out:

• The $u'$ terms cancel: $-2a + 2a = 0$
• The $u$ terms cancel: $a^2 - 2a^2 + a^2 = 0$

We are left with: $e^{-at}u'' = 0$

Step 4. Solve for $u(t)$: Since an exponential function $e^{-at}$ can never equal zero, we are left with the simple differential equation: $u''(t) = 0$

If the second derivative is zero, the first derivative must be a constant ($u'(t) = c_1$). Integrating again gives us a linear function: $u(t) = c_1t + c_2$

Step 5. Construct the second solution: Thus, $y_2(t) = e^{-at}(c_1t + c_2)$. Because the $c_2e^{-at}$ term is just a scalar multiple of our first solution $y_1(t)$, we discard it for the sake of linear independence. By setting $c_1 = 1$, we isolate the second independent solution: $y_2(t) = te^{-at}$

Conclusion: The general solution to the critically damped ODE is the linear combination of the two independent solutions we have just proven: $y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{-at} + c_2te^{-at}$ where $c_1$ and $c_2$ are arbitrary constants uniquely determined by the initial conditions (initial position and initial velocity).

Physical Significance and Interpretation

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 mathematics perfectly describes this physical reality $\boxed{y(t) = c_1y_1(t) + c_2y_2(t) = c_1e^{-at} + c_2te^{-at} = e^{-at}(c_1+c_2t)}$:

• The Exponential Decay: The term $e^{-at}$ drives the system exponentially toward zero (equilibrium).
• The Linear Term ($c_1+c_2t$): This term accommodates the system’s initial velocity. As time $t$ grows, the exponential decay $e^{-at}$ overpowers the linear growth of $t$. Because a linear function $c_1 + c_2t$ can only equal zero exactly once, the mathematical model guarantees the system can cross the equilibrium line at most one time before settling. It will never bounce back and forth (there is no oscillations). $y(t) = \underbrace{(c_1 + c_2 t)}{\text{straight line}} \times \underbrace{e^{-at}}{\text{shrinking}}$.
  A linear function: is a straight line, can cross zero at most once, and cannot wiggle, bend, or oscillate.

The critically damped response typically shows a smooth return to equilibrium without crossing the equilibrium position (no overshoot). It is the highly desired target state in mechanical and electrical systems designed to completely avoid structural oscillations. Common examples include car suspension systems (so the car doesn’t bounce after hitting a bump), automatic door closers (so the door shuts quickly without slamming or swinging open again), and precision analog dials.

Overdamped solutions look like: $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}, \quad r_1 < r_2 < 0.$ Two exponentials with different decay rates fight each other. The slower one makes the system sluggish. Critical damping replaces this with: $(c_1 + c_2 t)e^{-at}$ which decays faster than the slow exponential in the overdamped case.

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.