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PAGE 1

Fundamental Solutions of 2nd-Order Linear Eqs

Initial Value Problem Definition

\[ y'' + p(t)y' + q(t)y = g(t), \qquad y(t_0) = y_0, \quad y'(t_0) = y'_0 \]

\( p(t), q(t), g(t) \) do not contain \( y \) or \( y' \)

Linearity Condition: In a linear ordinary differential equation, the dependent variable \( y \) and all its derivatives (here \( y' \) and \( y'' \)) must appear linearly to the first power with no products or transcendental functions of \( y \). Thus, the coefficients \( p(t) \), \( q(t) \), and forcing term \( g(t) \) must be functions of the independent variable \( t \) only.

how do we know it has a solution? is it unique?

Existence & Uniqueness Theorem: If the coefficient functions \( p(t) \), \( q(t) \), and \( g(t) \) are continuous on an open interval \( I = (\alpha, \beta) \) containing \( t_0 \), then there exists a unique solution \( y = \phi(t) \) satisfying the initial value problem for all \( t \in I \).

notice the equation is an extension of

\[ y' + q(t)y = g(t) \quad \text{(pretend no } y''\text{)} \]

so, \( q(t), g(t) \) have to be continuous (from 1st-order)

First-Order Analogy: In first-order linear differential equations \( y' + p(t)y = g(t) \), continuity of the coefficient functions on an interval guarantees that the integrating factor \( e^{\int p(t)\,dt} \) exists and yields a unique solution.

Change of Variable

let's rewrite the 2nd-order eq. above w/ a change of variable

let \( x = y' \)

\[ x' + p(t)x = \underbrace{g(t) - q(t)y}_{\substack{\text{a function} \\ \text{of } t}} \]

← pretend we know \( y \) as function of \( t \)

Reduction of Order Concept: Setting \( x = y' \) means \( x' = y'' \). Substituting this into the second-order ODE yields \( x' + p(t)x = g(t) - q(t)y \). Treating \( y(t) \) as known temporarily transforms the left side into a first-order linear equation in \( x \).
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Existence and Uniqueness for Second-Order Linear Equations

it's another 1st-order linear, so we know \( p(t) \) has to be continuous

So, that means

\[ y'' + p(t)y' + q(t)y = g(t), \quad y(t_0) = y_0, \quad y'(t_0) = y_0' \]

has a unique solution on an interval containing \( t_0 \) on which ALL of \( p(t) \), \( q(t) \), \( g(t) \) are continuous.

Concept Note: Theorem Conditions

To apply the Existence and Uniqueness Theorem for linear differential equations, the equation must be written in standard form (where the coefficient of \( y'' \) is \( 1 \)). The theorem guarantees a unique solution on the largest open interval containing the initial point \( t_0 \) where all three functions—\( p(t) \), \( q(t) \), and \( g(t) \)—are simultaneously continuous.

Example

for example,

\[ t(t-4)y'' + 3ty' + 4y = \cos(t), \quad y(1) = 2, \quad y'(1) = 0 \]

Rewriting into standard form by dividing by \( t(t-4) \):

\[ y'' + \underbrace{\frac{3t}{t(t-4)}}_{p(t)} y' + \underbrace{\frac{4}{t(t-4)}}_{q(t)} y = \underbrace{\frac{\cos(t)}{t(t-4)}}_{g(t)} \]
Intervals of continuity for coefficient functions and the non-homogeneous term
Function Condition Intervals of Continuity
\( p(t) \) continuous on : \( (-\infty, 0) \), \( (0, 4) \), \( (4, \infty) \)
\( q(t) \) continuous on : \( (-\infty, 0) \), \( (0, 4) \), \( (4, \infty) \)
\( g(t) \) continuous on : \( (-\infty, 0) \), \( (0, 4) \), \( (4, \infty) \)

the interval containing \( t_0 = 1 \) where ALL are continuous is \( (0, 4) \) \( \longrightarrow \) there is a unique solution on this interval

Step-by-Step Breakdown

  1. Find Singularities: The denominator vanishes when \( t(t-4) = 0 \), meaning \( t = 0 \) and \( t = 4 \). Note that even if \( t \) cancels algebraically in \( p(t) \), the original ODE has a singular point at \( t = 0 \).
  2. Disjoint Intervals: The real line is partitioned into three open intervals: \( (-\infty, 0) \), \( (0, 4) \), and \( (4, \infty) \).
  3. Locate Initial Value: The initial conditions are specified at \( t_0 = 1 \). Since \( 1 \in (0, 4) \), the guaranteed interval of existence and uniqueness is \( (0, 4) \).
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Fundamental Solutions & Initial Conditions

We know if \( y_1 \) and \( y_2 \) are solutions (fundamental solutions) of

\[ y'' + p(t)y' + q(t)y = g(t) \]

then \( y = c_1 y_1 + c_2 y_2 \) is also a solution.

Note: In standard linear differential equation theory, superposition holds directly for the homogeneous equation (where the right-hand side is \( 0 \)). If non-homogeneous, \( c_1 y_1 + c_2 y_2 \) represents the complementary/homogeneous solution \( y_c(t) \).

where \( c_1 \) and \( c_2 \) depend on initial conditions \( y(t_0) = y_0 \), \( y'(t_0) = y_0' \).

Question: Can we always find \( c_1, c_2 \) for a given set of initial conditions?

Is that set of \( c_1 \) and \( c_2 \) unique?

(In other words, for given initial conditions, is there one and only one solution curve?)

\[ \begin{aligned} y &= c_1 y_1 + c_2 y_2 \\ y' &= c_1 y_1' + c_2 y_2' \end{aligned} \] \[ \begin{aligned} y(t_0) = y_0 &\longrightarrow c_1 y_1(t_0) + c_2 y_2(t_0) = y_0 \\ y'(t_0) = y_0' &\longrightarrow c_1 y_1'(t_0) + c_2 y_2'(t_0) = y_0' \end{aligned} \]
This system forms a linear system of two algebraic equations in two unknowns, \( c_1 \) and \( c_2 \). By Cramer's Rule, a unique pair \( (c_1, c_2) \) exists if and only if the determinant of coefficients (the Wronskian \( W(y_1, y_2)(t_0) = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \)) is nonzero.
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Matrix Form and the Wronskian

Put into matrix form:

Context: For a second-order linear initial value problem, the general solution \( y(t) = c_1 y_1(t) + c_2 y_2(t) \) must satisfy initial conditions \( y(t_0) = y_0 \) and \( y'(t_0) = y_0' \):

\[ \begin{aligned} c_1 y_1(t_0) + c_2 y_2(t_0) &= y_0 \\ c_1 y_1'(t_0) + c_2 y_2'(t_0) &= y_0' \end{aligned} \]

In matrix notation, this is written as \( A \mathbf{c} = \mathbf{y}_0 \).

\[ \underbrace{\begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{bmatrix}}_{\substack{\text{unique } c_1, c_2 \text{ exist} \\ \text{if this matrix is invertible}}} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} y_0 \\ y_0' \end{bmatrix} \]
\[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{bmatrix}^{-1} \begin{bmatrix} y_0 \\ y_0' \end{bmatrix} \]

The Invertibility Condition

It is invertible; its determinant must be nonzero:

\[ \underbrace{\begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix}}_{\text{this is called the Wronskian}} \neq 0 \]

Determinant & Invertibility: A square matrix is invertible if and only if \( \det(A) \neq 0 \). For a \( 2 \times 2 \) matrix, this guarantees a unique solution vector \( \mathbf{c} = A^{-1} \mathbf{y}_0 \).

\[ W[y_1, y_2](t_0) = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix} \]

if Wronskian \( \neq 0 \) then

we can ALWAYS find

\( c_1, c_2 \)

The Wronskian: The determinant \( W[y_1, y_2](t) = y_1(t) y_2'(t) - y_1'(t) y_2(t) \) characterizes the linear independence of solutions \( y_1 \) and \( y_2 \). If \( W \neq 0 \) at \( t_0 \), they form a fundamental set of solutions.

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Example: Wronskian & Linear Independence

For example, \( y'' + 8y' - 9y = 0 \) , with initial conditions \( y(t_0) = y_0 \), \( y'(t_0) = y_0' \) .

Step 1: Characteristic Equation: For a constant-coefficient homogeneous linear differential equation \( ay'' + by' + cy = 0 \), assume solutions of the form \( y = e^{rt} \). Substituting leads to the algebraic characteristic equation \( ar^2 + br + c = 0 \).

\[ r^2 + 8r - 9 = 0 \implies (r + 9)(r - 1) = 0 \implies r = -9, \; r = 1 \]

The two fundamental solutions are:

\[ y_1 = e^{-9t}, \qquad y_2 = e^t \]

Computing the Wronskian

The Wronskian Determinant: The Wronskian of two differentiable functions \( y_1 \) and \( y_2 \) is defined as \( W(y_1, y_2)(t) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2 \). If \( W(t_0) \neq 0 \), the functions form a fundamental set of solutions.

\[ \begin{aligned} W &= \begin{vmatrix} y_1(t_0) & y_2(t_0) \\[4pt] y_1'(t_0) & y_2'(t_0) \end{vmatrix} = \begin{vmatrix} e^{-9t_0} & e^{t_0} \\[4pt] -9e^{-9t_0} & e^{t_0} \end{vmatrix} \\[8pt] &= (e^{-9t_0})(e^{t_0}) - (-9e^{-9t_0})(e^{t_0}) \\[8pt] &= e^{-8t_0} + 9e^{-8t_0} = 10e^{-8t_0} \neq 0 \quad \text{for any } t_0 \end{aligned} \]

Conclusion:

  • So, \( C_1 \) and \( C_2 \) always exist.
  • This also means \( y_1 \) and \( y_2 \) are linearly independent.
  • (They span the entire solution space)
PAGE 6

Finding the Wronskian without Known Solutions

Question: Given \( y'' + p(t)y' + q(t)y = g(t) \),

Do we need \( y_1 \) and \( y_2 \) to find the Wronskian?

Surprisingly, no!

Context (Abel's Identity): This begins the derivation of Abel's Theorem, which demonstrates that the Wronskian \( W(y_1, y_2)(t) \) of two solutions satisfies a first-order separable differential equation dependent only on the coefficient \( p(t) \), not on the individual solutions \( y_1 \) and \( y_2 \).

Focus on the case with \( g(t) = 0 \):

\[ y'' + py' + qy = 0 \]

\( y_1 \) and \( y_2 \) are solutions,

so,

\[ \begin{aligned} y_1'' + py_1' + qy_1 &= 0 && \text{--- ①} \\[6pt] y_2'' + py_2' + qy_2 &= 0 && \text{--- ②} \end{aligned} \]

Multiply ① by \( -y_2 \) and ② by \( y_1 \):

\[ \begin{aligned} -y_2 y_1'' - y_2 p y_1' - y_2 q y_1 &= 0 \\[6pt] y_1 y_2'' + y_1 p y_2' + y_1 q y_2 &= 0 \end{aligned} \]

Next Step Preview: Adding these two equations cancels the \( q(t) \) terms because \( -y_2 q y_1 + y_1 q y_2 = 0 \). The remaining terms yield \( (y_1 y_2'' - y_2 y_1'') + p(y_1 y_2' - y_2 y_1') = 0 \), which is precisely \( W' + p(t)W = 0 \).

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Derivation of Abel's Identity

add them

\[ \overbrace{(-y_2 y_1'' + y_1 y_2'')}^{W'} + p\,\overbrace{(-y_2 y_1' + y_1 y_2')}^{W} = 0 \]
Context: Combining the two second-order linear differential equations for solutions \( y_1 \) and \( y_2 \) eliminates the zero-order term, leaving terms directly proportional to the Wronskian \( W \) and its derivative \( W' \).

remember

\[ W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_1' y_2 \]
\[ \begin{aligned} W' &= y_1 y_2'' + y_1' y_2' - (y_1' y_2' + y_1'' y_2) \\[6pt] &= y_1 y_2'' - y_2 y_1'' \end{aligned} \]
Derivative of the Wronskian: Differentiating \( W = y_1 y_2' - y_1' y_2 \) using the product rule produces intermediate cross terms \( y_1' y_2' \) that cancel out, leaving \( W' = y_1 y_2'' - y_2 y_1'' \).

the Wronskian satisfies the eq. \( W' + pW = 0 \)

1st-order linear and separable
\[ \begin{aligned} W' &= -pW \\[6pt] \frac{1}{W}\,dW &= -p\,dt \\[6pt] \ln |W| &= -pt + C \end{aligned} \]
\[ W = C e^{-\int p\,dt} \]

Abel's formula / identity

(allows finding Wronskian w/o \( y_1, y_2 \))

Significance of Abel's Identity: Abel's formula proves that the Wronskian \( W(t) \) is either identically zero or never zero on an interval where \( p(t) \) is continuous, completely bypassing the need to solve explicitly for \( y_1(t) \) and \( y_2(t) \).
PAGE 8

Example: Abel's Formula for the Wronskian

For example,

\[ y'' - \frac{2}{t}y' + \frac{2}{t^2}y = 0 \]

In standard form \( y'' + p(t)y' + q(t)y = 0 \), the coefficient of \( y' \) gives \( p(t) \).

Identifying \( p = -\frac{2}{t} \), we compute the Wronskian \( W \) using Abel's formula:

\[ \begin{aligned} W &= C e^{-\int p \, dt} \\[6pt] &= C e^{-\int -\frac{2}{t} \, dt} \\[6pt] &= \underbrace{C}_{\substack{\text{any } C \neq 0}} t^2 \end{aligned} \]

Integrating \( -\int -\frac{2}{t} \, dt = 2\ln|t| = \ln(t^2) \), which gives \( e^{\ln(t^2)} = t^2 \).

So, choosing \( C = 1 \):

Result:
\[ W = t^2 \]