Second-Order Linear ODEs

A second-order linear ODE has the form

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

Theorem 3.1. If $y_1$ and $y_2$ are solutions of the homogeneous equation $y'' + py' + qy = 0$ Then $c_1 y_1 + c_2 y_2$ is also a solution (superposition principle).

Theorem 3.2 (Wronskian Criterion). Two solutions $y_1, y_2$ of the homogeneous equation form a fundamental set (i.e., span all solutions) if and only if their Wronskian is non-zero:

$W(y_1, y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \neq 0$

Abel’s identity states that $W(x) = W(x_0) e^{-\int_{x_0}^x p(t)\, dt}$.

Theorem 3.3. The general solution of $y'' + py' + qy = g$ is $y = y_h + y_p$Where $y_h$ is the General solution of the homogeneous equation and $y_p$ is any particular solution.

For $y'' + ay' + by = 0$ with $a, b$ constants, try $y = e^{rx}$:

$r^2 + ar + b = 0$

Case 1: Two distinct real roots $r_1 \neq r_2$. $y_h = c_1 e^{r_1 x} + c_2 e^{r_2 x}$.

Case 2: Repeated root $r$. $y_h = c_1 e^{rx} + c_2 x e^{rx}$.

Case 3: Complex conjugate roots $r = \alpha \pm i\beta$. $y_h = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))$.

Problem. Solve $y'' - 5y' + 6y = 0$ with $y(0) = 1$, $y'(0) = 0$.

Solution. Characteristic equation: $r^2 - 5r + 6 = (r-2)(r-3) = 0$. Roots: $r = 2, 3$.

$y_h = c_1 e^{2x} + c_2 e^{3x}$.

$y(0) = c_1 + c_2 = 1$. $y'(0) = 2c_1 + 3c_2 = 0$. Solving: $c_1 = 3$, $c_2 = -2$.

$y = 3e^{2x} - 2e^{3x}$. $\blacksquare$

Problem. Solve $y'' + 2y' + 5y = 0$ with $y(0) = 1$, $y'(0) = 0$.

Problem. Solve $y'' - 4y' + 4y = 0$ with $y(0) = 1$, $y'(0) = 3$.

For equations $y'' + ay' + by = g(x)$ where $g(x)$ is a polynomial, exponential, sine, cosine, or Products of these, guess the form of $y_p$ and solve for coefficients.

Rule. If any term of the guess is a solution of the homogeneous equation, multiply by $x$ (or $x^2$ if already multiplied by $x$).

Problem. Solve $y'' - y = 2e^x$.

Solution. Homogeneous: $r^2 - 1 = 0$Roots $\pm 1$. $y_h = c_1 e^x + c_2 e^{-x}$.

Since $e^x$ is a homogeneous solution, guess $y_p = Axe^x$. $y_p' = Ae^x + Axe^x$ $y_p'' = 2Ae^x + Axe^x$. $y_p'' - y_p = (2Ae^x + Axe^x) - Axe^x = 2Ae^x = 2e^x$So $A = 1$.

$y = c_1 e^x + c_2 e^{-x} + xe^x$. $\blacksquare$

Problem. Solve $y'' + 3y' + 2y = x^2 + 1$.

Problem. Solve $y'' + 2y' + y = 3e^{-x}\sin x$.

Consider the forced harmonic oscillator

$y'' + \omega_0^2 y = F_0 \cos(\omega t)$

Case 1: $\omega \neq \omega_0$ (Non-resonant). The particular solution is $y_p = \frac{F_0}{\omega_0^2 - \omega^2}\cos(\omega t)$With bounded amplitude.

Case 2: $\omega = \omega_0$ (Resonant). Since $\cos(\omega_0 t)$ is a homogeneous solution, Guess $y_p = At\sin(\omega_0 t)$. Substituting:

$y_p'' + \omega_0^2 y_p = 2A\omega_0 \cos(\omega_0 t) - A\omega_0^2 t\sin(\omega_0 t) + A\omega_0^2 t\sin(\omega_0 t) = 2A\omega_0 \cos(\omega_0 t)$

Setting equal to $F_0 \cos(\omega_0 t)$: $A = \frac{F_0}{2\omega_0}$.

$y_p = \frac{F_0}{2\omega_0} t \sin(\omega_0 t)$

The amplitude grows linearly with $t$ --- this is resonance. Physically, the system absorbs energy From the periodic forcing at its natural frequency, causing unbounded oscillations.

Worked Example. Solve $y'' + 9y = 6\cos(3t)$, $y(0) = 0$, $y'(0) = 0$.

Theorem 3.4 (Variation of Parameters). For $y'' + p(x)y' + q(x)y = g(x)$Let $y_1, y_2$ be a Fundamental set of solutions of the homogeneous equation. Then a particular solution is

$y_p = -y_1 \int \frac{y_2 g}{W}\, dx + y_2 \int \frac{y_1 g}{W}\, dx$

Where $W = W(y_1, y_2) = y_1 y_2' - y_2 y_1'$.

Proof. Seek $y_p = u_1(x)y_1(x) + u_2(x)y_2(x)$. Impose the constraint $u_1'y_1 + u_2'y_2 = 0$. Then $y_p' = u_1 y_1' + u_2 y_2'$ and $y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''$. Substituting into the ODE: $(u_1'y_1' + u_2'y_2') + u_1(y_1'' + py_1' + qy_1) + u_2(y_2'' + py_2' + qy_2) = g$. Since $y_1, y_2$ satisfy the homogeneous equation, this reduces to $u_1'y_1' + u_2'y_2' = g$. Together With $u_1'y_1 + u_2'y_2 = 0$Solving gives the formulas above. $\blacksquare$

Problem. Solve $y'' + y = \tan x$ using variation of parameters.

Solution. $y_1 = \cos x$, $y_2 = \sin x$. $W = \cos x \cdot \cos x - \sin x \cdot (-\sin x) = 1$.

$u_1' = -\frac{y_2 g}{W} = -\sin x \tan x = -\frac{\sin^2 x}{\cos x} = -(1 - \cos^2 x)/\cos x = -\sec x + \cos x$.

$u_1 = -\ln|\sec x + \tan x| + \sin x$.

$u_2' = \frac{y_1 g}{W} = \cos x \tan x = \sin x$.

$u_2 = -\cos x$.

$y_p = (-\ln|\sec x + \tan x| + \sin x)\cos x + (-\cos x)\sin x = -\cos x \ln|\sec x + \tan x|$.

$y = c_1 \cos x + c_2 \sin x - \cos x \ln|\sec x + \tan x|$. $\blacksquare$

Theorem 3.5. Given one solution $y_1(x)$ of $y'' + p(x)y' + q(x)y = 0$A second linearly Independent solution is obtained by setting $y_2 = y_1 \int \frac{e^{-\int p(x)\, dx}}{y_1^2}\, dx$.

Proof. Seek $y_2 = v(x) y_1(x)$. Then $y_2' = v'y_1 + vy_1'$ and $y_2'' = v''y_1 + 2v'y_1' + vy_1''$. Substituting into the ODE:

$v''y_1 + 2v'y_1' + vy_1'' + p(v'y_1 + vy_1') + qvy_1 = 0$

$v''y_1 + v'(2y_1' + py_1) + v(y_1'' + py_1' + qy_1) = 0$

Since $y_1$ satisfies the ODE, the coefficient of $v$ vanishes:

$v''y_1 + v'(2y_1' + py_1) = 0$

Let $w = v'$. Then $w'y_1 + w(2y_1' + py_1) = 0$A separable first-order ODE:

$\frac{w'}{w} = -\frac{2y_1' + py_1}{y_1} = -2\frac{y_1'}{y_1} - p$

$\ln w = -2\ln y_1 - \int p\, dx \implies w = \frac{e^{-\int p\, dx}}{y_1^2}$

Since $w = v'$We obtain the result. $\blacksquare$

Worked Example. Given that $y_1 = e^x$ solves $y'' - 2y' + y = 0$Find a second solution.

An Euler-Cauchy (equidimensional) equation has the form

$x^2 y'' + axy' + by = 0, \quad x > 0$

The substitution $y = x^r$ gives the characteristic equation

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

Case 1: Two distinct real roots $r_1 \neq r_2$. $y_h = c_1 x^{r_1} + c_2 x^{r_2}$.

Case 2: Repeated root $r$. $y_h = c_1 x^r + c_2 x^r \ln x$.

Case 3: Complex roots $r = \alpha \pm i\beta$. $y_h = x^{\alpha}(c_1 \cos(\beta \ln x) + c_2 \sin(\beta \ln x))$.

Problem. Solve $x^2 y'' - 3xy' + 4y = 0$.

Worked Example. Solve $x^2 y'' + xy' + y = 0$.

For $y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1 y' + a_0 y = 0$:

• Characteristic equation $r^n + a_{n-1}r^{n-1} + \cdots + a_0 = 0$.
• For root $r$ of multiplicity $m$: include $e^{rx}, xe^{rx}, \ldots, x^{m-1}e^{rx}$.
• For complex roots $\alpha \pm i\beta$ of multiplicity $m$: include $e^{\alpha x} x^k \cos(\beta x)$ and $e^{\alpha x} x^k \sin(\beta x)$ for $k = 0, \ldots, m - 1$.

A mass $m$ on a spring with spring constant $k$ and damping coefficient $c$Subject to external force $F(t)$Satisfies

$mx'' + cx' + kx = F(t)$

Dividing by $m$ and setting $\omega_0 = \sqrt{k/m}$, $\gamma = c/(2m)$:

$x'' + 2\gamma x' + \omega_0^2 x = \frac{F(t)}{m}$

The homogeneous solution depends on the discriminant $\gamma^2 - \omega_0^2$:

The underdamped oscillation $e^{-\gamma t}\cos(\omega_d t)$ (blue) decays inside the envelope $\pm e^{-\gamma t}$ (green/red). Adjust sliders c (damping $\gamma$) and w (damped frequency $\omega_d$) to explore different regimes.

:::caution Common Pitfall When using undetermined coefficients, always check whether your guess Overlaps with the homogeneous solution. For $y'' - 4y = e^{2x}$Guessing $y_p = Ae^{2x}$ fails Because $e^{2x}$ satisfies the homogeneous equation. You must use $y_p = Axe^{2x}$ instead. :::

:::caution Common Pitfall For Euler-Cauchy equations, the substitution $y = x^r$ only works for $x > 0$. For $x < 0$Substitute $x = -e^t$ or use $y = (-x)^r$. :::

:::caution Common Pitfall Variation of parameters always works but can lead to difficult integrals. If the forcing term $g(x)$ is a polynomial, exponential, sine, or cosine (or products of these), Prefer undetermined coefficients --- it is much faster.

Theorem 3.6 (Abel’s Identity). If $y_1, y_2$ are solutions of $y'' + p(x)y' + q(x)y = 0$ Then their Wronskian satisfies

$W(x) = W(x_0) e^{-\int_{x_0}^x p(t)\, dt}$

Proof. Since $y_1, y_2$ satisfy the ODE:

$y_1'' = -py_1' - qy_1$ and $y_2'' = -py_2' - qy_2$.

$W' = y_1 y_2'' + y_1' y_2' - y_1'' y_2 - y_1' y_2'$

$= y_1(-py_2' - qy_2) - (-py_1' - qy_1)y_2$

$= -p(y_1 y_2' - y_1' y_2) = -pW$.

So $W' + pW = 0$Giving $W = Ce^{-\int p\, dx}$And evaluating at $x_0$ gives the result. $\blacksquare$

Corollary. $W(x)$ is either identically zero or never zero.

Problem. Solve $y'' - 4y = xe^x$ using variation of parameters.

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