Second-Order Ordinary Differential Equations (ODEs)

A second-order linear differential equation has the form

$$ \frac{d^2y}{dt^2} + p(t)\frac{dy}{dt} + q(t)y = g(t) $$

where $ g(t) $ is called the forcing or source term, and determines how the system is driven externally.

The equation is homogeneous if $ g(t) = 0 \; \forall \; t \in I, $ and inhomogeneous otherwise.

A function $ \phi(t) $ is a solution on $ I $ if $ \phi(t) $, $ \phi'(t) $, and $ \phi''(t) $ exist on $ I $, and

$$ \phi'' + p\phi' + q\phi = g $$

for all $ t \in I $. An initial value problem (IVP) specifies both the value and slope at some time $t_0$:

$$ y(t_0) = y_0, \quad y'(t_0) = y_0'$$

Superposition Principle

As with first-order linear equations, second-order linear equations satisfy a superposition property. Suppose $ y_1 $ and $ y_2 $ satisfy

$$ \frac{d^2y_1}{dt^2} + p(t)\frac{dy_1}{dt} + q(t)y_1 = g_1(t) $$ $$ \frac{d^2y_2}{dt^2} + p(t)\frac{dy_2}{dt} + q(t)y_2 = g_2(t) $$

Then the linear combination $ y = c_1 y_1 + c_2 y_2 $ satisfies

$$ \frac{d^2y}{dt^2} + p(t)\frac{dy}{dt} + q(t)y = c_1 g_1(t) + c_2 g_2(t) $$

This reflects the linear structure of the equation: combining solutions produces a corresponding combination on the right-hand side.

Now let $ y_p(t) $ be a particular solution of the original equation, and let $ y(t) $ be any solution. Then

$$ (y - y_p)'' + p(t)(y - y_p)' + q(t)(y - y_p) = g - g = 0 $$

so $ y - y_p $ satisfies the homogeneous equation

$$ \frac{d^2y}{dt^2} + p(t)\frac{dy}{dt} + q(t)y = 0 $$

Therefore,

$$ y(t) - y_p(t) = y_c(t) $$

and the general solution can be written as

$$ y(t) = y_p(t) + y_c(t) $$

where $ y_c(t) $ is the complementary solution, the general solution of the homogeneous equation.

Homogeneous Equation and Linear Combinations

Suppose $ y_1 $ and $ y_2 $ satisfy the homogeneous equation

$$ y_1'' + py_1' + qy_1 = 0 $$ $$ y_2'' + py_2' + qy_2 = 0 $$

Then any linear combination $ c_1 y_1 + c_2 y_2 $ also satisfies the equation. This follows directly by substitution:

$$ (c_1 y_1 + c_2 y_2)'' + p(c_1 y_1 + c_2 y_2)' + q(c_1 y_1 + c_2 y_2) = 0 $$

The important question is whether we can always choose $ c_1 $ and $ c_2 $ to satisfy given initial conditions.

The initial conditions require

$$ c_1 y_1(t_0) + c_2 y_2(t_0) = y_0 $$ $$ c_1 y_1'(t_0) + c_2 y_2'(t_0) = y_0' $$

This leads to the matrix equation

$$ \begin{pmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} y_0 \\ y_0' \end{pmatrix} $$

which is of the form $ A\vec{c} = \vec{y}_0. $ The vector $ \vec{c} $ is uniquely defined if $ A^{-1} $ exists, and this happens $ \Leftrightarrow \det(A) \neq 0 $. So a unique solution exists if and only if the determinant of this matrix is nonzero.

Wronskian Determinant

The determinant

$$ W(y_1,y_2)(t) = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t) \end{vmatrix} $$

is called the Wronskian, and measures whether two solutions are linearly independent.

$$ W(y_1,y_2)(t) = y_1(t)y_2'(t) - y_2(t)y_1'(t) $$

The condition $ W(y_1,y_2)(t_0) \neq 0 $ ensures that any initial conditions can be satisfied.

In this case, $ y_1 $ and $ y_2 $ form a fundamental set of solutions, and

$$ y_c(t) = c_1 y_1(t) + c_2 y_2(t) $$

gives the general solution of the homogeneous equation.

Generality of Fundamental Solutions

Suppose $ y_1(t) $ and $ y_2(t) $ are solutions of the homogeneous equation with $ W(y_1,y_2)(t_0) \neq 0. $ We show that this condition holds for all $ t \in I $.

$$ y_1'' + py_1' + qy_1 = 0 $$ $$ y_2'' + py_2' + qy_2 = 0 $$

Multiply the first equation by $ y_2 $ and the second by $ y_1 $, then subtract:

$$ y_1 y_2'' + p y_1 y_2' + q y_1 y_2 - y_2 y_1'' - p y_2 y_1' - q y_2 y_1 = 0 $$

The $ q $-terms cancel, giving

$$ y_1 y_2'' - y_2 y_1'' + p(y_1 y_2' - y_2 y_1') = 0 $$ $$ \frac{d}{dt}(y_1 y_2' - y_2 y_1') + p(t)(y_1 y_2' - y_2 y_1') = 0 $$

Let $ W(t) = y_1 y_2' - y_2 y_1'. $ Then

$$ \frac{dW}{dt} + p(t)W = 0 $$

This is a first-order linear equation, whose solution is

$$ W(t) = W(t_0) \, e^{-\int_{t_0}^{t} p(s)\,ds} $$

Therefore,

$$ W(t) \neq 0 \Leftrightarrow W(t_0) \neq 0 $$

So if two solutions are independent at one point, they are independent everywhere.

Linear Dependence of Functions

Two functions $ f $ and $ g $ are said to be linearly dependent on the interval $I$ if there exist constants $ k_1 $ and $ k_2 $, not both zero, such that

$$ k_1 f(t) + k_2 g(t) = 0 $$

for all $ t \in I $. The Wronskian provides a convenient test:

If $ W(f,g)(t_0) \neq 0 $ for some $ t_0 \in I $, then $ f $ and $ g $ are linearly independent on $ I $.

If $ f $ and $ g $ are linearly dependent on $ I $, then $ W(f,g)(t) = 0 $ on $ I $.

eg. Consider $ \color{#00f0ff} t^2 y'' - 2y = 0. $

We will do this in three parts.

Part 1. Show that this equation has two solutions of the form $ t^n $.

Suppose $ y = ct^n $ is a solution. Then

$$ y' = c n t^{n-1}, \quad y'' = c n (n-1) t^{n-2} $$

Substituting into the differential equation gives

$$ t^2 \left(cn(n-1)t^{n-2}\right) - 2ct^n = 0 $$ $$ c\left(n^2 - n - 2\right)t^n = 0 $$

If $ c \neq 0 $ and $ t \neq 0 $, then

$$ n^2 - n - 2 = 0 $$ $$ (n+1)(n-2) = 0 $$ $$ n = 2, \quad n = -1 $$

Hence two solutions are

$$ y_1(t) = t^{-1}, \quad y_2(t) = t^2 $$

Part 2. Show that $ y_1(t) $ and $ y_2(t) $ are linearly independent.

Their Wronskian is

$$ W(y_1,y_2)(t) = \begin{vmatrix} t^{-1} & t^2 \\ -t^{-2} & 2t \end{vmatrix} $$ $$ = 2t^{-1}t + t^2t^{-2} = 3 \ne 0 $$

Therefore $ y_1 $ and $ y_2 $ are linearly independent.

Part 3. Find the solution satisfying $ \, y(1) = 0, \, y'(1) = -3 $

The general solution is

$$ y(t) = c_1 t^{-1} + c_2 t^2 $$

Differentiate

$$ y'(t) = -c_1 t^{-2} + 2c_2 t $$ $$ y(1) = 0 = c_1 + c_2 \quad \Rightarrow \quad c_1 = -c_2 $$ $$ y'(1) = -3 = -c_1 + 2c_2 $$ $$ \Rightarrow \, c_2 = -1, \quad c_1 = 1 $$ $$ \therefore y(t) = t^{-1} - t^2 $$