Second-Order Linear Differential Equations, Mechanical Vibrations, and Variable Coefficient Methods

Fundamentals of Second-Order Linear Differential Equations

Second-order differential equations play a central role in mathematics and physics because they model physical systems governed by acceleration, such as Newton's second law of motion, $m y''(t) = F(t)$. The general form of a second-order linear differential equation is given by:

$a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x)$

where $a_2(x) \neq 0$ on the interval of interest. When the forcing function $f(x)$ is identically zero, the equation is called homogeneous:

$a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$

Dividing through by the leading coefficient function $a_2(x)$ yields the standard homogeneous form:

$y'' + p(x)y' + q(x)y = 0$

where $p(x) = \frac{a_1(x)}{a_2(x)}$ and $q(x) = \frac{a_0(x)}{a_2(x)}$.

While a first-order differential equation such as $y' = ky$ requires only a single initial condition $y(0) = y_0$ to uniquely determine an arbitrary constant, a second-order linear differential equation requires two initial conditions to determine a unique specific solution:

$y(0) = y_0$

$y'(0) = v_0$

Geometrically and physically, these conditions specify the initial position and the initial velocity of the state variable.

The fundamental algebraic structure of the solution space for homogeneous linear differential equations is governed by the Principle of Superposition. Let $y_1(x)$ and $y_2(x)$ be any two solutions of the homogeneous equation $a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$. Then any linear combination:

$y(x) = C_1 y_1(x) + C_2 y_2(x)$

is also a solution of the differential equation for arbitrary constants $C_1$ and $C_2$.

This principle can be verified directly by substitution. Substituting $y(x) = C_1 y_1(x) + C_2 y_2(x)$ into the differential operator yields:

$a_2(x)(C_1 y_1 + C_2 y_2)'' + a_1(x)(C_1 y_1 + C_2 y_2)' + a_0(x)(C_1 y_1 + C_2 y_2)$

$= C_1 [a_2(x)y_1'' + a_1(x)y_1' + a_0(x)y_1] + C_2 [a_2(x)y_2'' + a_1(x)y_2' + a_0(x)y_2]$

Because $y_1$ and $y_2$ are solutions, both bracketed expressions equal zero:

$= C_1(0) + C_2(0) = 0$

To construct the general solution to a second-order homogeneous linear equation, the two solutions must be linearly independent. Not every pair of solutions suffices; for instance, the functions $e^x$ and $2e^x$ are constant multiples of each other, meaning they span only a one-dimensional solution space. Conversely, for the equation $y'' - y = 0$, the two functions $y_1(x) = e^x$ and $y_2(x) = e^{-x}$ are linearly independent because neither is a scalar multiple of the other. Thus, the general solution of $y'' - y = 0$ is:

$y(x) = C_1 e^x + C_2 e^{-x}$

Homogeneous Equations with Constant Coefficients

When the coefficient functions are real constants, the equation takes the form:

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

where $a \neq 0$, and $b$ and $c$ are real constants. Because differentiating exponential functions reproduces the original function scaled by a constant factor, a trial solution of the form $y = e^{rx}$ is utilized. Differentiating yields $y' = r e^{rx}$ and $y'' = r^2 e^{rx}$. Substituting these derivatives into the differential equation gives:

$a(r^2 e^{rx}) + b(r e^{rx}) + c(e^{rx}) = 0$

$(a r^2 + b r + c)e^{rx} = 0$

Since $e^{rx} \neq 0$ for all real and complex values of $x$, the polynomial expression in parentheses must vanish. This leads to the characteristic equation (also known as the auxiliary equation):

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

The nature of the roots of this quadratic polynomial determines the functional form of the general solution. There are three mutually exclusive cases based on the discriminant $\Delta = b^2 - 4ac$.

In Case 1, the discriminant is strictly positive ($b^2 - 4ac > 0$), yielding two distinct real roots $r_1 \neq r_2$. In this scenario, the two solutions $y_1(x) = e^{r_1 x}$ and $y_2(x) = e^{r_2 x}$ are linearly independent. The general solution is:

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

Consider the differential equation $y'' - 5y' + 6y = 0$. Its characteristic equation is:

$r^2 - 5r + 6 = 0$

$(r - 2)(r - 3) = 0$

This gives the distinct real roots $r_1 = 2$ and $r_2 = 3$. The general solution is:

$y(x) = C_1 e^{2x} + C_2 e^{3x}$

To solve the initial value problem with initial conditions $y(0) = 1$ and $y'(0) = 0$, first evaluate $y(0)$:

$y(0) = C_1 e^0 + C_2 e^0 = C_1 + C_2 = 1$

Next, compute the derivative of the general solution:

$y'(x) = 2 C_1 e^{2x} + 3 C_2 e^{3x}$

Evaluating the derivative at $x = 0$ yields:

$y'(0) = 2 C_1 + 3 C_2 = 0$

From the first equation, $C_1 = 1 - C_2$. Substituting this into the velocity equation gives:

$2(1 - C_2) + 3 C_2 = 0$

$2 - 2 C_2 + 3 C_2 = 0$

$C_2 = -2$

$C_1 = 1 - (-2) = 3$

Thus, the specific solution to the initial value problem is:

$y(x) = 3 e^{2x} - 2 e^{3x}$

In Case 2, the discriminant is zero ($b^2 - 4ac = 0$), yielding a single repeated real root $r = -\frac{b}{2a}$. The exponential guess yields only one independent solution, $y_1(x) = e^{rx}$. A second linearly independent solution is formed by multiplying by $x$, giving $y_2(x) = x e^{rx}$. The general solution is:

$y(x) = (C_1 + C_2 x)e^{rx}$

Consider the equation $y'' + 6y' + 9y = 0$ subject to initial conditions $y(0) = 2$ and $y'(0) = -4$. The characteristic equation is:

$r^2 + 6r + 9 = 0$

$(r + 3)^2 = 0$

This yields the repeated root $r = -3$. The general solution is:

$y(x) = (C_1 + C_2 x)e^{-3x}$

Applying the first initial condition $y(0) = 2$:

$y(0) = (C_1 + C_2(0))e^0 = C_1 = 2$

Differentiating the general solution via the product rule:

$y'(x) = C_2 e^{-3x} - 3(C_1 + C_2 x)e^{-3x}$

Applying the second initial condition $y'(0) = -4$:

$y'(0) = C_2 - 3 C_1 = -4$

$C_2 - 3(2) = -4$

$C_2 - 6 = -4$

$C_2 = 2$

Therefore, the specific solution to the initial value problem is:

$y(x) = (2 + 2x)e^{-3x}$

In Case 3, the discriminant is strictly negative ($b^2 - 4ac < 0$), yielding complex conjugate roots of the form $r = \omega \pm i \epsilon$, where $\epsilon \neq 0$. Using Euler's formula, $e^{i \theta} = \cos(\theta) + i \sin(\theta)$, the complex exponential solutions can be transformed into real-valued basis functions:

$e^{(\omega \pm i \epsilon)x} = e^{\omega x} e^{\pm i \epsilon x} = e^{\omega x}(\cos(\epsilon x) \pm i \sin(\epsilon x))$

Taking linear combinations to extract the real and imaginary components yields the general real solution:

$y(x) = e^{\omega x}(C_1 \cos(\epsilon x) + C_2 \sin(\epsilon x))$

Consider the undamped harmonic oscillator $y'' + 4y = 0$ with $y(0) = 0$ and $y'(0) = 6$. The characteristic equation is:

$r^2 + 4 = 0$

$r = \pm 2i$

Here, the real part is $\omega = 0$ and the imaginary frequency parameter is $\epsilon = 2$. The general solution is:

$y(x) = C_1 \cos(2x) + C_2 \sin(2x)$

Applying the initial position condition $y(0) = 0$:

$y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 = 0$

Differentiating the remaining term $y(x) = C_2 \sin(2x)$:

$y'(x) = 2 C_2 \cos(2x)$

Applying the initial velocity condition $y'(0) = 6$:

$y'(0) = 2 C_2 = 6$

$C_2 = 3$

Thus, the solution is purely oscillatory with constant amplitude:

$y(x) = 3 \sin(2x)$

Now consider an equation exhibiting oscillation with decay: $y'' + 2y' + 5y = 0$. The characteristic equation is:

$r^2 + 2r + 5 = 0$

$r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i$

Here, $\omega = -1$ and $\epsilon = 2$. The general solution is:

$y(x) = e^{-x}(C_1 \cos(2x) + C_2 \sin(2x))$

Unlike the pure oscillation of $y'' + 4y = 0$, the presence of the negative real part $\omega = -1$ introduces the damping envelope $e^{-x}$, causing the oscillations to attenuate toward zero as $x \rightarrow \infty$.

A quick classification of characteristic roots is illustrated by comparing $y'' - 9y = 0$ and $y'' + 6y' + 9y = 0$. In the former, $r^2 - 9 = 0 \implies r = \pm 3$, which corresponds to Case 1 (two distinct real roots). In the latter, $r^2 + 6r + 9 = (r + 3)^2 = 0 \implies r = -3$, which corresponds to Case 2 (a repeated real root).

Mathematical Modeling of Mass-Spring Systems

Consider a body of mass $m$ (measured in $\text{kg}$) attached to a spring moving horizontally on a surface. Let $x(t)$ represent the signed displacement of the mass from its static equilibrium position at time $t$, where displacement to the right is defined as positive ($x > 0$) and displacement to the left is negative ($x < 0$). Static equilibrium corresponds to $x = 0$.

According to Hooke's Law, when a spring is stretched or compressed, it exerts a restoring force $F_s$ proportional to the displacement from equilibrium:

$F_s = -k x$

where $k > 0$ is the spring constant, measured in $\text{N/m}$. The negative sign indicates that the restoring force always opposes the direction of displacement.

When resistive forces are present, such as viscous fluid drag or mechanical friction, they are modeled to first approximation as proportional to velocity:

$F_d = -c x'$

where $c > 0$ is the damping coefficient, measured in $\text{N}\cdot\text{s/m}$. The negative sign reflects that friction opposes the instantaneous direction of motion.

Let $F_{\text{ext}}(t)$ denote an externally applied driving force measured in Newtons ($\text{N}$). By Newton's second law, the equation of motion is $m x'' = \sum F$:

$m x'' = F_s + F_d + F_{\text{ext}}(t)$

$m x'' = -k x - c x' + F_{\text{ext}}(t)$

Rearranging gives the second-order linear differential equation governing mechanical vibrations:

$m x'' + c x' + k x = F_{\text{ext}}(t)$

When no driving force acts on the system ($F_{\text{ext}}(t) = 0$), the motion is classified as free motion:

$m x'' + c x' + k x = 0$

When multiple springs are combined, their equivalent stiffness depends on their physical arrangement. For two springs arranged in parallel with spring constants $k_1$ and $k_2$, both springs experience the exact same displacement $x$. The net restoring force is the sum of the individual forces:

$F = F_1 + F_2 = -k_1 x - k_2 x = -(k_1 + k_2)x$

Therefore, the effective spring constant for springs in parallel is:

$k_{\text{eff}} = k_1 + k_2$

For two springs arranged in series, both springs transmit the exact same internal force magnitude $F$, while their individual displacements $x_1$ and $x_2$ add to produce the total displacement $x = x_1 + x_2$. Since $x_1 = -\frac{F}{k_1}$ and $x_2 = -\frac{F}{k_2}$:

$x = -F\left(\frac{1}{k_1} + \frac{1}{k_2}\right)$

Solving for the force $F$ gives:

$F = -\left(\frac{1}{k_1} + \frac{1}{k_2}\right)^{-1} x$

Thus, the effective spring constant for two springs in series is:

$k_{\text{eff}} = \left(\frac{1}{k_1} + \frac{1}{k_2}\right)^{-1} = \frac{k_1 k_2}{k_1 + k_2}$

In a vertical mass-spring system, gravity acts continuously downward. When a mass $m$ is attached to an unstretched vertical spring, gravity causes an initial static elongation $\Delta$ until the system reaches static equilibrium, satisfying:

$k \Delta = m g$

Let $x(t)$ represent the additional dynamic displacement measured downward from this static equilibrium position. The total elongation of the spring is $x + \Delta$. The net downward force acting on the mass is:

$F_{\text{net}} = m g - k(x + \Delta) = m g - k x - k \Delta$

Because $k \Delta = m g$, the gravitational force cancels exactly with the static spring pre-tension:

$F_{\text{net}} = m g - k x - m g = -k x$

Applying Newton's second law shows that the dynamic equation of motion for a vertical spring, when measured from static equilibrium, is identical to that of the horizontal system:

$m x'' + c x' + k x = F_{\text{ext}}(t)$

Dynamics of Free Undamped Vibrations

When damping is negligible ($c = 0$) and no external driving force acts ($F_{\text{ext}} = 0$), the governing equation reduces to the undamped free oscillation model:

$m x'' + k x = 0$

Dividing by the mass $m$ yields:

$x'' + \frac{k}{m} x = 0$

The characteristic equation is $m r^2 + k = 0$, which gives the purely imaginary roots:

$r = \pm i \sqrt{\frac{k}{m}}$

The parameter $\omega_0 = \sqrt{\frac{k}{m}}$ is defined as the natural angular frequency of the system, measured in $\text{rad/s}$. The general solution can be written as a linear combination of trigonometric functions:

$x(t) = C_1 \cos(\omega_0 t) + C_2 \sin(\omega_0 t)$

This linear combination can be expressed as a single phase-shifted cosine wave:

$x(t) = R \cos(\omega_0 t - \delta)$

Using the cosine angle-difference identity, this expression expands to:

$R \cos(\omega_0 t - \delta) = R \cos(\delta)\cos(\omega_0 t) + R \sin(\delta)\sin(\omega_0 t)$

Matching coefficients with the original form gives:

$C_1 = R \cos(\delta)$

$C_2 = R \sin(\delta)$

Squaring and adding these equations yields the oscillation amplitude $R$:

$R^2 = C_1^2 + C_2^2$

$R = \sqrt{C_1^2 + C_2^2}$

Dividing the second equation by the first yields the phase angle $\delta$:

$\tan(\delta) = \frac{C_2}{C_1}$

A function $u(t)$ is periodic if there exists a positive number $T > 0$ such that $u(t + T) = u(t)$ for all $t$. The smallest such value $T$ is the period. For $x(t) = R \cos(\omega_0 t - \delta)$, periodicity requires:

$R \cos(\omega_0(t + T) - \delta) = R \cos(\omega_0 t - \delta + \omega_0 T) = R \cos(\omega_0 t - \delta)$

This equality holds when $\omega_0 T = 2\pi$. Consequently, the period $T$ and natural cyclic frequency $f$ are:

$T = \frac{2\pi}{\omega_0}$

$f = \frac{1}{T} = \frac{\omega_0}{2\pi}$

As a concrete example, consider an object of mass $m = 2\,\text{kg}$ attached to a spring on a frictionless horizontal surface with spring constant $k = 8\,\text{N/m}$. The initial displacement is $x(0) = 0.1\,\text{m}$ and the initial velocity is $x'(0) = 0.2\,\text{m/s}$. The governing differential equation is:

$2 x'' + 8 x = 0$

$x'' + 4 x = 0$

The natural angular frequency is:

$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{8}{2}} = 2\,\text{rad/s}$

The general solution is:

$x(t) = C_1 \cos(2t) + C_2 \sin(2t)$

Applying the initial condition $x(0) = 0.1\,\text{m}$:

$x(0) = C_1 = 0.1$

Differentiating to obtain velocity:

$x'(t) = -2 C_1 \sin(2t) + 2 C_2 \cos(2t)$

Applying $x'(0) = 0.2\,\text{m/s}$:

$x'(0) = 2 C_2 = 0.2$

$C_2 = 0.1$

Thus, the specific equation of motion is:

$x(t) = 0.1 \cos(2t) + 0.1 \sin(2t)$

In a second scenario, a $0.1\,\text{kg}$ mass suspended vertically stretches a spring by $0.15\,\text{m}$ under gravity ($g = 9.8\,\text{m/s}^2$). Using static equilibrium $k \Delta = m g$:

$k = \frac{m g}{\Delta} = \frac{0.1 \times 9.8}{0.15} = \frac{0.98}{0.15} = \frac{98}{15} \approx 6.533\,\text{N/m}$

If this mass and spring are placed on a frictionless horizontal surface in a vacuum, pulled $0.3\,\text{m}$ to the right, and released with an initial leftward speed of $2\,\text{m/s}$, the angular frequency is:

$\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{98/15}{0.1}} = \sqrt{\frac{980}{15}} = \sqrt{\frac{196}{3}} \approx 8.083\,\text{rad/s}$

Approximating $\omega_0 \approx 8\,\text{rad/s}$, the general solution is:

$x(t) = C_1 \cos(8t) + C_2 \sin(8t)$

The initial conditions are $x(0) = 0.3$ and $x'(0) = -2$ (the negative sign indicates leftward velocity). Substituting $x(0)$

$x(0) = C_1 = 0.3$

Differentiating gives:

$x'(t) = -8 C_1 \sin(8t) + 8 C_2 \cos(8t)$

$x'(0) = 8 C_2 = -2$

$C_2 = -0.25$

The resulting equation of motion is:

$x(t) = 0.3 \cos(8t) - 0.25 \sin(8t)$

Free Damped Vibrations and Damping Regimes

When viscous damping is included in free motion ($F_{\text{ext}} = 0$), the equation of motion is:

$m x'' + c x' + k x = 0$

with $m > 0$, $c > 0$, and $k > 0$. The characteristic equation is:

$m r^2 + c r + k = 0$

Applying the quadratic formula yields the roots:

$r = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}$

The sign of the discriminant $\Delta = c^2 - 4mk$ defines three distinct physical damping regimes.

Underdamped motion occurs when $c^2 < 4mk$. In this case, the discriminant is negative, producing complex conjugate roots:

$r = -\gamma \pm i \omega_d$

where the exponential decay parameter $\gamma$ and the damped natural frequency (quasi-frequency) $\omega_d$ are defined as:

$\gamma = \frac{c}{2m}$

$\omega_d = \sqrt{\frac{k}{m} - \left(\frac{c}{2m}\right)^2} = \sqrt{\omega_0^2 - \gamma^2}$

The general solution is:

$x(t) = e^{-\gamma t}(C_1 \cos(\omega_d t) + C_2 \sin(\omega_d t)) = R e^{-\gamma t}\cos(\omega_d t - \delta)$

The motion consists of rapid oscillations at frequency $\omega_d$ (which is strictly less than the natural frequency $\omega_0$) trapped inside an exponentially decaying envelope defined by $\pm R e^{-\gamma t}$.

Critical damping occurs when $c^2 = 4mk$. The discriminant vanishes, yielding a single repeated real root:

$r = -\frac{c}{2m}$

The general solution is:

$x(t) = (C_1 + C_2 t)e^{-\frac{c}{2m}t}$

Under critical damping, the system returns to equilibrium as rapidly as possible without sustained oscillations. The trajectory can pass through equilibrium ($x = 0$) at most once before decaying asymptotically to zero. Critical damping represents the operational threshold between oscillatory and purely dissipative behavior.

Overdamped motion occurs when $c^2 > 4mk$. The discriminant is strictly positive, yielding two distinct, negative real roots $r_1, r_2 < 0$:

$r_1 = \frac{-c + \sqrt{c^2 - 4mk}}{2m}$

$r_2 = \frac{-c - \sqrt{c^2 - 4mk}}{2m}$

The general solution is a sum of purely decaying exponentials:

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

Because the damping force dominates, the mass moves sluggishly back toward equilibrium without oscillating.

To illustrate underdamped motion, consider a system with mass $m = 1\,\text{kg}$, spring constant $k = 25\,\text{N/m}$, and damping coefficient $c = 6\,\text{N}\cdot\text{s/m}$. The initial displacement is $x(0) = 0.10\,\text{m}$ and the system is released from rest ($x'(0) = 0\,\text{m/s}$). Evaluating the discriminant:

$c^2 - 4mk = 6^2 - 4(1)(25) = 36 - 100 = -64 < 0$

Because the discriminant is negative, the system is underdamped. The characteristic roots are:

$r = \frac{-6 \pm \sqrt{-64}}{2(1)} = -3 \pm 4i$

Here, $\gamma = 3$ and $\omega_d = 4\,\text{rad/s}$. The general solution is:

$x(t) = e^{-3t}(C_1 \cos(4t) + C_2 \sin(4t))$

Applying $x(0) = 0.10$:

$x(0) = C_1 = 0.10$

Differentiating $x(t)$ via the product rule:

$x'(t) = -3 e^{-3t}(C_1 \cos(4t) + C_2 \sin(4t)) + e^{-3t}(-4 C_1 \sin(4t) + 4 C_2 \cos(4t))$

Evaluating at $t = 0$:

$x'(0) = -3 C_1 + 4 C_2 = 0$

$-3(0.10) + 4 C_2 = 0$

$4 C_2 = 0.3$

$C_2 = 0.075$

Thus, the specific equation of motion is:

$x(t) = e^{-3t}(0.10 \cos(4t) + 0.075 \sin(4t))$

The influence of changing the damping coefficient can be examined for fixed parameters $m = 1\,\text{kg}$ and $k = 9\,\text{N/m}$, with fixed initial conditions $x(0) = 1$ and $x'(0) = 0$. The critical damping value is:

$c_{\text{crit}} = \sqrt{4mk} = \sqrt{4(1)(9)} = 6\,\text{N}\cdot\text{s/m}$

For $c = 1\,\text{N}\cdot\text{s/m}$, $c^2 = 1 < 36$, so the system is underdamped and oscillates with decaying amplitude. For $c = 6\,\text{N}\cdot\text{s/m}$, $c^2 = 36$, so the system is critically damped and achieves the fastest non-oscillatory return to equilibrium. For $c = 8\,\text{N}\cdot\text{s/m}$, $c^2 = 64 > 36$, so the system is overdamped, resulting in an exponential decay to equilibrium that is slower than in the critically damped case.

Euler-Cauchy Equations

When a second-order linear differential equation has variable coefficients, standard exponential trial functions $e^{rx}$ generally fail. However, a significant class of variable-coefficient equations can be solved algebraically: the Euler-Cauchy (or equidimensional) equations. The general homogeneous second-order Euler-Cauchy equation is:

$x^2 y'' + a x y' + b y = 0$

for $x > 0$, where $a$ and $b$ are constants. Notice that the power of $x$ in each coefficient matches the order of the derivative it multiplies: $x^2 y''$, $x^1 y'$, and $x^0 y$. Higher-order equations follow this pattern, such as $a_3 x^3 y''' + a_2 x^2 y'' + a_1 x y' + a_0 y = 0$.

Because each differentiation of a power function $x^m$ lowers the power by one, multiplying by the corresponding power of $x$ restores the exponent to $m$. Thus, a power function trial solution is employed:

$y = x^m$

Differentiating gives:

$y' = m x^{m-1}$

$y'' = m(m - 1)x^{m-2}$

Substituting these derivatives into the differential equation gives:

$x^2 [m(m - 1)x^{m-2}] + a x [m x^{m-1}] + b [x^m] = 0$

$m(m - 1)x^m + a m x^m + b x^m = 0$

$[m(m - 1) + a m + b]x^m = 0$

Since $x > 0$, $x^m \neq 0$, yielding the Euler-Cauchy auxiliary equation:

$m(m - 1) + a m + b = 0$

$m^2 + (a - 1)m + b = 0$

This quadratic equation produces three distinct cases depending on its roots.

In Case 1, the auxiliary equation has two distinct real roots $m_1 \neq m_2$. The general solution is:

$y(x) = C_1 x^{m_1} + C_2 x^{m_2}$

For example, consider the equation $x^2 y'' - 2x y' + 2y = 0$ for $x > 0$. Here $a = -2$ and $b = 2$. The auxiliary equation is:

$m^2 + (-2 - 1)m + 2 = 0$

$m^2 - 3m + 2 = 0$

$(m - 1)(m - 2) = 0$

This gives distinct roots $m_1 = 1$ and $m_2 = 2$. The general solution is:

$y(x) = C_1 x + C_2 x^2$

To solve the initial value problem with $y(1) = 3$ and $y'(1) = 4$, evaluate $y(1)$:

$y(1) = C_1(1) + C_2(1)^2 = C_1 + C_2 = 3$

Computing the derivative:

$y'(x) = C_1 + 2 C_2 x$

Evaluating at $x = 1$:

$y'(1) = C_1 + 2 C_2 = 4$

Subtracting the first equation from the second:

$(C_1 + 2 C_2) - (C_1 + C_2) = 4 - 3$

$C_2 = 1$

$C_1 = 3 - 1 = 2$

Thus, the specific solution is:

$y(x) = 2x + x^2$

In Case 2, the auxiliary equation has a repeated real root $m$. The power trial solution yields $y_1(x) = x^m$. A second linearly independent solution is $y_2(x) = x^m \ln(x)$. The general solution is:

$y(x) = x^m(C_1 + C_2 \ln(x))$

Consider the equation $x^2 y'' - 3x y' + 4y = 0$. Here $a = -3$ and $b = 4$. The auxiliary equation is:

$m^2 + (-3 - 1)m + 4 = 0$

$m^2 - 4m + 4 = 0$

$(m - 2)^2 = 0$

This gives a double root at $m = 2$. The general solution is:

$y(x) = x^2(C_1 + C_2 \ln(x))$

In Case 3, the auxiliary equation has complex conjugate roots $m = \omega \pm i \epsilon$, where $\epsilon \neq 0$. Using the identity $x^m = e^{m \ln(x)}$:

$x^{\omega \pm i \epsilon} = e^{(\omega \pm i \epsilon)\ln(x)} = e^{\omega \ln(x)} e^{\pm i \epsilon \ln(x)} = x^\omega [\cos(\epsilon \ln(x)) \pm i \sin(\epsilon \ln(x))]$

Extracting linearly independent real solutions gives:

$y(x) = x^\omega [C_1 \cos(\epsilon \ln(x)) + C_2 \sin(\epsilon \ln(x))]$

Consider $x^2 y'' - x y' + 5y = 0$. Here $a = -1$ and $b = 5$. The auxiliary equation is:

$m^2 + (-1 - 1)m + 5 = 0$

$m^2 - 2m + 5 = 0$

$m = \frac{2 \pm \sqrt{4 - 20}}{2} = 1 \pm 2i$

Here $\omega = 1$ and $\epsilon = 2$. The general solution for $x > 0$ is:

$y(x) = x [C_1 \cos(2 \ln(x)) + C_2 \sin(2 \ln(x))]$

When solving Euler-Cauchy equations on the domain $x < 0$, the variable $x$ is replaced by its absolute value $|x|$, yielding basis terms such as $|x|^m$ and $\ln|x|$.

The Method of Reduction of Order

For a general second-order linear homogeneous differential equation with variable coefficients:

$y'' + p(x)y' + q(x)y = 0$

knowing one non-zero solution $y_1(x)$ allows a second linearly independent solution $y_2(x)$ to be determined using the method of reduction of order.

To understand the underlying mechanism, consider the special case where $q(x) = 0$:

$y'' + p(x)y' = 0$

Because the dependent variable $y$ does not appear explicitly, $y_1(x) = 1$ is an obvious solution. Substituting $w(x) = y'(x)$ reduces the second-order equation to a first-order linear differential equation in $w$:

$w' + p(x)w = 0$

Separating variables yields:

$\frac{dw}{w} = -p(x)dx$

$\ln|w| = -\int p(x)dx$

$w(x) = w(x_0)e^{-\int p(x)dx}$

Because $w(x) = y'(x)$, integrating once more recovers the second solution:

$y_2(x) = \int w(x)dx = \int w(x_0)e^{-\int p(x)dx}dx$

For the general case where $q(x) \neq 0$, a second solution is sought in the form of a variable multiple of the known solution:

$y_2(x) = u(x)y_1(x)$

Differentiating using the product rule:

$y_2'(x) = u' y_1 + u y_1'$

$y_2''(x) = u'' y_1 + 2 u' y_1' + u y_1''$

Substituting $y_2$, $y_2'$, and $y_2''$ into $y'' + p(x)y' + q(x)y = 0$:

$(u'' y_1 + 2 u' y_1' + u y_1'') + p(x)(u' y_1 + u y_1') + q(x)(u y_1) = 0$

Regrouping terms by the derivatives of $u(x)$:

$y_1 u'' + [2 y_1' + p(x)y_1]u' + [y_1'' + p(x)y_1' + q(x)y_1]u = 0$

Because $y_1$ is an exact solution of the original differential equation, the coefficient multiplying $u$ vanishes identically:

$y_1'' + p(x)y_1' + q(x)y_1 = 0$

This simplifies the equation to:

$y_1 u'' + [2 y_1' + p(x)y_1]u' = 0$

Setting $v(x) = u'(x)$ reduces this second-order equation for $u$ into a first-order separable differential equation for $v$: