ODE: Second-Order Homogeneous with Constant Coefficients

Mastering second-order linear homogeneous differential equations with constant coefficients is a cornerstone of engineering analysis. These equations are the fundamental mathematical models for a vast array of physical systems, from the vibrations of a car suspension and the oscillations of an RLC circuit to the buckling of structural columns. Learning to solve them efficiently unlocks your ability to predict and control dynamic system behavior.

The Characteristic Equation: The Key to the Solution

The standard form for the equation we are solving is:

$$y'' + b y' + c y = 0$$

where $y$ is the unknown function of $t$ (or $x$), and $b$ and $c$ are real-valued constants. The characteristic equation is the brilliant algebraic shortcut that converts this calculus problem into an algebra problem. We derive it by proposing an exponential trial solution of the form $y = e^{rt}$, where $r$ is an unknown constant. Substituting into our ODE:

$$y = e^{rt}, \quad y' = r e^{rt}, \quad y'' = r^2 e^{rt}$$

Plugging these into $y'' + b y' + c y = 0$ gives:

$$r^2 e^{rt} + b r e^{rt} + c e^{rt} = 0$$

Since $e^{rt}$ is never zero, we can factor it out, leaving the characteristic equation:

$$r^2 + b r + c = 0$$

This is a simple quadratic equation. The nature of its roots—distinct real, complex conjugate, or repeated—directly determines the mathematical form of the general solution to our original differential equation.

Case 1: Distinct Real Roots

When the discriminant $b^2 - 4c > 0$, the characteristic equation yields two distinct real roots, $r_1$ and $r_2$. Each root corresponds to a valid exponential solution: $e^{r_1 t}$ and $e^{r_2 t}$. Because our ODE is linear and homogeneous, any linear combination of these individual solutions is also a solution. Therefore, the general solution is:

$$y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$$

where $C_1$ and $C_2$ are arbitrary constants determined by initial conditions.

Physical Interpretation: This case models overdamped systems. Think of a door closing on a strong hydraulic closer. When opened and released, it returns to its closed position without oscillating, simply decaying back smoothly. The two distinct real roots (both negative for stable systems) represent two different rates of exponential decay.

Worked Example: Solve $y'' + 5y' + 6y = 0$.

1. Write the characteristic equation: $r^2 + 5r + 6 = 0$.
2. Solve: $(r + 2)(r + 3) = 0$, so $r_1 = -2$, $r_2 = -3$.
3. General solution: $y(t) = C_1 e^{-2t} + C_2 e^{-3t}$.

Case 2: Complex Conjugate Roots

When the discriminant $b^2 - 4c < 0$, the roots are complex conjugates: $r = \alpha \pm \beta i$, where $\alpha = -b/2$ and $\beta = \sqrt{4c - b^2}/2$. The fundamental solutions are $e^{(\alpha + \beta i)t}$ and $e^{(\alpha - \beta i)t}$. Using Euler's formula ($e^{i\theta} = \cos \theta + i \sin \theta$), we can convert these into real-valued functions. The resulting general solution is:

$$y(t) = e^{\alpha t} \big( C_1 \cos(\beta t) + C_2 \sin(\beta t) \big)$$

An equivalent, often useful form is $y(t) = A e^{\alpha t} \cos(\beta t - \phi)$, where $A$ is the amplitude and $\phi$ is the phase shift.

Physical Interpretation: This case models underdamped oscillatory systems. A classic example is a mass on a spring moving through air. When displaced, it oscillates back and forth with a frequency $\beta$ (the damped natural frequency), while the amplitude decays exponentially at a rate determined by $\alpha$. If $\alpha = 0$ (i.e., $b=0$), the system is undamped and oscillates forever.

Worked Example: Solve $y'' + 2y' + 5y = 0$.

1. Characteristic equation: $r^2 + 2r + 5 = 0$.
2. Roots: $r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$. So, $\alpha = -1$, $\beta = 2$.
3. General solution: $y(t) = e^{-t} ( C_1 \cos(2t) + C_2 \sin(2t) )$.

Case 3: Repeated Root

When the discriminant $b^2 - 4c = 0$, there is a single, repeated real root $r = -b/2$. We have one obvious solution: $y_1 = e^{rt}$. To find a second, linearly independent solution, we use the method of reduction of order, which leads us to the solution $y_2 = t e^{rt}$. The general solution is:

$$y(t) = C_1 e^{rt} + C_2 t e^{rt} = e^{rt}(C_1 + C_2 t)$$

Physical Interpretation: This special case represents critical damping. It is the precise damping level that separates oscillatory from non-oscillatory motion. A critically damped system, like a high-quality car shock absorber or the needle on a high-end analog meter, returns to equilibrium in the shortest possible time without oscillating. Engineers often tune systems to this point for the fastest non-oscillatory response.

Worked Example: Solve $y'' + 6y' + 9y = 0$.

1. Characteristic equation: $r^2 + 6r + 9 = 0$.
2. Roots: $(r + 3)^2 = 0$, so $r = -3$ (repeated).
3. General solution: $y(t) = C_1 e^{-3t} + C_2 t e^{-3t}$.

Solving Initial Value Problems

An initial value problem (IVP) appends specific conditions, typically $y(0)$ and $y'(0)$, to the differential equation. The solution process is systematic:

1. Find the general solution $y(t)$ based on the root case.
2. Compute its derivative $y'(t)$.
3. Substitute $t=0$ into both $y(t)$ and $y'(t)$.
4. Set up and solve the resulting system of two equations for $C_1$ and $C_2$.

IVP Example (Complex Roots): Solve $y'' + 4y' + 13y = 0$ with $y(0)=2, y'(0)=1$.

1. Characteristic: $r^2 + 4r + 13=0$. Roots: $r = -2 \pm 3i$. So $\alpha=-2, \beta=3$.

General: $y(t) = e^{-2t}(C_1 \cos 3t + C_2 \sin 3t)$.

1. Derivative: $y'(t) = e^{-2t}[(-2C_1+3C_2)\cos 3t + (-2C_2-3C_1)\sin 3t]$.
2. Apply $y(0)=2$: $e^{0}(C_1 \cos 0 + C_2 \sin 0) = C_1 = 2$.
3. Apply $y'(0)=1$: $e^{0}[(-2(2)+3C_2)\cos 0 + (...)\sin 0] = -4 + 3C_2 = 1$. Thus, $3C_2 = 5$, so $C_2 = 5/3$.
4. Particular Solution: $y(t) = e^{-2t}(2 \cos 3t + \frac{5}{3} \sin 3t)$.

Common Pitfalls

1. Forgetting the $t$ in the repeated root case. The most frequent error is writing $y = C_1 e^{rt} + C_2 e^{rt}$ for a repeated root. This is not a general solution because the two functions are not linearly independent. You must include the factor of $t$ in the second term: $C_2 t e^{rt}$.

1. Incorrectly applying initial conditions to the complex form. When working with the solution $y(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t)$, you must apply initial conditions to this form, not to the intermediate complex exponential form. Differentiate carefully using the product rule.

1. Misidentifying the root case. Always calculate the discriminant ($b^2 - 4c$) first. This tells you which formula to use and prevents you from trying to take the square root of a negative number without introducing the imaginary unit $i$.

1. Arithmetic errors in the characteristic equation. The coefficients in the characteristic equation $r^2 + b r + c = 0$ come directly from the ODE $y'' + b y' + c y = 0$. A sign error here propagates through the entire solution. Double-check that $b$ and $c$ are correctly transferred, including their signs.

Summary

• The characteristic equation $r^2 + b r + c = 0$, derived from the trial solution $y=e^{rt}$, transforms the problem of solving $y'' + b y' + c y = 0$ into solving a quadratic.
• The three possible root cases dictate the general solution:
• Distinct Real Roots $(r_1, r_2)$: $y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$ (Overdamped).
• Complex Conjugate Roots $(\alpha \pm \beta i)$: $y(t) = e^{\alpha t} (C_1 \cos \beta t + C_2 \sin \beta t)$ (Underdamped/Oscillatory).
• Repeated Root $(r)$: $y(t) = e^{rt} (C_1 + C_2 t)$ (Critically Damped).
• Each case has a direct physical interpretation in engineering systems involving vibration, oscillation, or dynamic response.
• Solving an initial value problem requires finding the general solution, then using the given initial conditions $y(0)$ and $y'(0)$ to solve for the constants $C_1$ and $C_2$, yielding a unique, particular solution.