Second Order Linear ODEs

Standard Form

A second order linear ODE with constant coefficients:

$\frac{d^2y}{dx^2} + a\frac{dy}{dx} + by = f(x)$

or equivalently:

$y'' + 2\gamma y' + \omega_0^2 y = f(x)$

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Homogeneous Case ($f(x) = 0$)

Try solution $y = e^{\lambda x}$:

$\lambda^2 e^{\lambda x} + 2\gamma \lambda e^{\lambda x} + \omega_0^2 e^{\lambda x} = 0$

Dividing by $e^{\lambda x}$ gives the characteristic equation:

$\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0$

Roots: $\lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$

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Three Cases for the Roots

Case	Condition	Roots	General Solution
Two real distinct	$\gamma^2 > \omega_0^2$	$\lambda_1, \lambda_2$ (real)	$y = Ae^{\lambda_1 x} + Be^{\lambda_2 x}$
Repeated root	$\gamma^2 = \omega_0^2$	$\lambda = -\gamma$ (double)	$y = (A + Bx)e^{\lambda x}$
Complex conjugate	$\gamma^2 < \omega_0^2$	$-\gamma \pm i\Omega$	$y = e^{-\gamma x}(A\cos\Omega x + B\sin\Omega x)$

where $\Omega = \sqrt{\omega_0^2 - \gamma^2}$.

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Derivation of the Repeated Root Solution

When $\lambda_1 = \lambda_2$, we only have one solution $y_1 = e^{\lambda x}$.

Method of reduction of order: Seek second solution as $y_2 = v(x)e^{\lambda x}$.

Substituting and solving gives $v' =$ constant, so $v = Bx$.

Hence $y = (A + Bx)e^{\lambda x}$.

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Inhomogeneous Case: Particular Integral

Principle of Superposition

General solution = Complementary Function + Particular Integral

$y = y_c + y_p$

• $y_c$: general solution to homogeneous equation
• $y_p$: any particular solution to inhomogeneous equation

Finding Particular Integral

Use trial solutions based on the form of $f(x)$:

$f(x)$	Try $y_p$
Polynomial of degree $n$	General polynomial of degree $n$
$Fe^{Kx}$	$Ae^{Kx}$ (unless $K = \lambda_1$ or $\lambda_2$)
$F\sin Kx$ or $F\cos Kx$	$A\sin Kx + B\cos Kx$
Product/compound	Combine forms

If trial form appears in $y_c$: Multiply by $x$ (or $x^2$ for double root).

Examples

Example 1: $y'' + 5y' + 6y = 12$

Characteristic: $\lambda^2 + 5\lambda + 6 = 0 \Rightarrow \lambda = -2, -3$

$y_c = Ae^{-2x} + Be^{-3x}$

Try $y_p = C$. Then $y_p'' = 0$, $y_p' = 0$, so $6C = 12$, giving $C = 2$.

$y = Ae^{-2x} + Be^{-3x} + 2$

Example 2: $y'' + 4y = \cos x$

Characteristic: $\lambda^2 + 4 = 0 \Rightarrow \lambda = \pm 2i$

$y_c = A\cos 2x + B\sin 2x$

Try $y_p = C\cos x + D\sin x$. Substituting:

$(-C + 4C)\cos x + (-D + 4D)\sin x = \cos x$

$3C\cos x + 3D\sin x = \cos x$

$C = \frac{1}{3}, D = 0$

$y = A\cos 2x + B\sin 2x + \frac{1}{3}\cos x$

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Initial Value Problems

Apply initial conditions $y(x_0) = y_0$, $y'(x_0) = y_1$ to determine $A$ and $B$.

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Physical Applications

• Springs: $m\ddot{x} + c\dot{x} + kx = F(t)$
• Circuits: $L\ddot{q} + R\dot{q} + \frac{q}{C} = V(t)$
• Pendulum (small angle): $\ddot{\theta} + \frac{g}{L}\theta = 0$

The mathematical structure is universal; the physics provides the coefficients.