# Second-Order Linear ODEs

Physics I · Differential Equations and Vector Calculus · https://tryals.app/learn/physics-i/second-order-linear-odes

A **Second-Order Linear ODE** takes the form $ay'' + by' + cy = f(x)$. When the **homogeneous case** applies ($f=0$), we test $y = e^{rx}$ to find the characteristic equation $ar^2 + br + c = 0$.

The general solution depends on the discriminant $\Delta = b^2 - 4ac$:

| Case | Roots | General Solution ($y$) |
| --- | --- | --- |
| $\Delta > 0$ | $r_1 \neq r_2$ (real) | $C_1 e^{r_1 x} + C_2 e^{r_2 x}$ |
| $\Delta = 0$ | $r$ (repeated) | $(C_1 + C_2 x)e^{rx}$ |
| $\Delta < 0$ | $\alpha \pm \beta i$ (complex) | $e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$ |

For **non-homogeneous** equations, the complete solution is **$y = y_h + y_p$** using undetermined coefficients or variation of parameters.

## Damped Oscillators & Pitfalls

**Common pitfall:** Stopping at $y_h$ misses the steady-state ($y_p$), while stopping at $y_p$ drops the transient. You **must** include both.

Adding damping to a spring gives $m\ddot{x} + b\dot{x} + kx = 0$. Behaviour is governed by the **damping ratio** $\zeta = \frac{b}{2\sqrt{mk}}$:

| Damping Type | Ratio Condition | Physical Behaviour |
| --- | --- | --- |
| **Underdamped** | $\zeta < 1$ | Oscillations with exponential decay |
| **Critically damped** | $\zeta = 1$ | Fastest return to zero without oscillation |
| **Overdamped** | $\zeta > 1$ | Slow, sluggish exponential return |

**Physics link:** Critical damping ($\zeta = 1$) is used in door closers and shock absorbers to achieve equilibrium in minimum time.

## Practice questions

5 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. For $y'' + by' + ky = 0$ (a damped oscillator), the roots of the characteristic equation are **complex** with negative real part. Without solving anything else, what does the solution do?

A. It grows without bound
B. It oscillates forever with constant amplitude
C. It oscillates while decaying to zero
D. It decays smoothly to zero with no oscillation

**Answer:** C. It oscillates while decaying to zero

**Why:** Complex roots mean oscillation; a negative real part wraps it in a shrinking exponential envelope, underdamped ringing, like a struck bell. The root locations are a complete qualitative dictionary: no solution formula needed.

Page: https://tryals.app/practice/physics-i/second-order-linear-odes/for-y-by-ky-0-a-damped-oscillator-the-roots-of-the

### 2. The damped oscillator’s behaviour changes qualitatively as damping increases, passing through a critical value. Why does the behaviour change qualitatively at that value?

A. The amplitude reaches zero and the system stops moving entirely
B. The equation stops being linear and must be solved numerically
C. The frequency becomes infinite, which is why the motion ceases
D. The characteristic roots become real, so oscillation stops

**Answer:** D. The characteristic roots become real, so oscillation stops

**Why:** The discriminant changes sign and complex roots become real. This is why critical damping is the design target for a door closer or a car suspension: fastest return with no overshoot.

Page: https://tryals.app/practice/physics-i/second-order-linear-odes/the-damped-oscillators-behaviour-changes-qualitatively-as-damping

### 3. Order the steps for solving a second-order linear ODE with constant coefficients.

**Answer:**

1. Write the characteristic equation $r^{2}$ + br + $c = 0$
2. Compute the discriminant $b^{2}$ − 4c
3. Find the roots $r_{1}$, $r_{2}$
4. Write the general solution based on root type
5. Apply initial conditions to find $C_{1}$ and $C_{2}$

**Why:** This systematic approach works for any constant-coefficient 2nd-order ODE.

Page: https://tryals.app/practice/physics-i/second-order-linear-odes/order-the-steps-for-solving-a-second-order-linear-ode-with-constant

### 4. Sort each solution behaviour by the **roots** of its characteristic equation.

**Answer:**

- Two real negative roots: Smooth decay, no oscillation
- Complex roots, negative real part: Decaying oscillation, A struck tuning fork ringing down
- Pure imaginary roots: Steady oscillation, constant amplitude

**Why:** The complex plane is a behaviour map: left half = decay, right half = growth, off-axis = oscillation. Engineers judge the stability of aircraft and circuits by plotting the roots and checking which half-plane they occupy.

Page: https://tryals.app/practice/physics-i/second-order-linear-odes/sort-each-solution-behaviour-by-the-roots-of-its-characteristic

### 5. The general solution of a second-order ODE has two arbitrary constants, and a first-order one has a single constant. Why does the order fix the count?

A. Because the constants come in pairs for dimensional consistency
B. Because second-order equations have two solutions rather than one
C. Because each integration introduces one constant
D. Because a quadratic always yields exactly two solution pieces

**Answer:** C. Because each integration introduces one constant

**Why:** Order equals the number of conditions needed. Newton's second law is second order, which is exactly why initial position and velocity, and nothing more, determine a trajectory.

Page: https://tryals.app/practice/physics-i/second-order-linear-odes/the-general-solution-of-a-second-order-ode-has-two-arbitrary