4.1: General Theory of nth Order Linear Equations Thm: L(y) = y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y is a linear function. Proof: Let a, b be real numbers. L(af + bg) = (af + bg)^{(n)} + p_{1}(t)(af + bg)^{(n−1)} + ... + p_{n−1}(t)(af + bg)′ + p_{n}(t)(af + bg) = af^{(n)} + bg^{(n)} + p_{1}(af^{(n−1)} + bg^{(n−1)}) + ... + p_{n−1}(af′ + bg′) + p_{n}(af + bg) = af^{(n)}+p_{1}af^{(n−1)}+...+p_{n−1}af′+p_{n}af+bg^{(n)}+p_{1}bg^{(n−1)}+...+p_{n−1}bg′+p_{n}bg = a[f^{(n)}+p_{1}f^{(n−1)}+...+p_{n−1}f′+p_{n}f]+b[g^{(n)}+p_{1}g^{(n−1)}+...+p_{n−1}g′+p_{n}g]

= aL(f) + bL(g)

Theorem: If ϕ_{i}, i = 1, ..., n are solutions to a homogeneous linear differential equation (i.e., y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0 (*) ) , then c_{1}ϕ_{1} + c_{2}ϕ_{2} + ... + c_{n}ϕ_{n} is also a solution to this linear differential equation. Pf: Since ϕ_{i}, i = 1, ..., n are solutions to (*), L(ϕ_{i}) = 0 for i = 1, ..., n. Thus L(c_{1}ϕ_{1} + c_{2}ϕ_{2} + ... + c_{n}ϕ_{n}) = c_{1}L(ϕ_{1}) + c_{2}L(ϕ_{2}) + ... + c_{n}L(ϕ_{n}) = 0. Thus c_{1}ϕ_{1} + c_{2}ϕ_{2} + ... + c_{n}ϕ_{n} is also a solution to (*). Solve: y′′ + y = 0, y(0) = −1, y′(0) = −3 r^{2} + 1 = 0 implies r^{2} = −1. Thus r = ±i. Since r = 0 ± 1i, y = k_{1}cos(t) + k_{2}sin(t). Then y′ = −k_{1}sin(t) + k_{2}cos(t) y(0) = −1: −1 = k_{1}cos(0) + k_{2}sin(0) implies −1 = k_{1} y′(0) = −3: −3 = −k_{1}sin(0) + k_{2}cos(0) implies −3 = k_{2} Thus IVP solution: y = −cos(t) − 3sin(t) When does the following IVP have a unique solution: IVP: y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0,

y(t_{0}) = y_{0}, y′(t_{0}) = y_{1}, ..., y^{(n−1)}(t_{0}) = y_{n−1}.

Suppose y = c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) + ... + c_{n}ϕ_{n}(t) is a solution to this IVT Then y(t_{0}) = y_{0}: y_{0} = c_{1}ϕ_{1}(t_{0}) + c_{2}ϕ_{2}(t_{0}) + ... + c_{n}ϕ_{n}(t_{0})

y′(t_{0}) = y_{1}: y_{1} = c_{1}ϕ′_{1}(t_{0}) + c_{2}ϕ′_{2}(t_{0}) + ... + c_{n}ϕ′_{n}(t_{0})

. . .

y^{(n−1)}(t_{0}) = y_{n−1}: y_{n−1} = c_{1}ϕ_{1}^{(n−1)}(t_{0}) + c_{2}ϕ_{2}^{(n−1)}(t_{0}) + ... + c_{n}ϕ_{n}^{(n−1)}(t_{0})

To find IVP solution, need to solve above system of equations for the unknowns c_{i}, i = 1, ..., n.

Note the IVP has a unique solution if and only if the above system of equations has a unique solution for the c_{i}.

Note that in these equations the c_{i} are the unknowns and the y_{i}, ϕ_{i}(t_{0}), ...ϕ_{i}^{(n−1)}(t_{0}), are the constants. We can translate this linear system of equations into matrix form:

[ \begin{matrix}ϕ_{1}(t_{0}) & ϕ_{2}(t_{0}) & ... & ϕ_{n}(t_{0}) \\ ϕ′_{1}(t_{0}) & ϕ′_{2}(t_{0}) & ... & ϕ′_{n}(t_{0}) \\ & . \\ & . \\ & . \\ ϕ_{1}^{(n−1)}(t_{0}) & ϕ_{2}^{(n−1)}(t_{0}) & ... & ϕ_{n}^{(n−1)}(t_{0}) \end{matrix} ] [ \begin{matrix}c_{1} \\ c_{2} \\ . \\ . \\ . \\ c_{n} \end{matrix} ] = [ \begin{matrix}y_{0} \\ y_{1} \\ . \\ . \\ . \\ y_{n−1} \end{matrix} ]

Note this equation has a unique solution if and only if

det [ \begin{matrix}ϕ_{1}(t_{0}) & ϕ_{2}(t_{0}) & ... & ϕ_{n}(t_{0}) \\ ϕ′_{1}(t_{0}) & ϕ′_{2}(t_{0}) & ... & ϕ′_{n}(t_{0}) \\ & . \\ & . \\ & . \\ ϕ_{1}^{(n−1)}(t_{0}) & ϕ_{2}^{(n−1)}(t_{0}) & ... & ϕ_{n}^{(n−1)}(t_{0}) \end{matrix} ] ≠ 0

Definition: The Wronskian of the functions, ϕ_{1}, ϕ_{2},..., ϕ_{n} is

W (ϕ_{1}, ϕ_{2}, ..., ϕ_{n}) = det [ \begin{matrix}ϕ_{1}(t) & ϕ_{2}(t) & ... & ϕ_{n}(t) \\ ϕ′_{1}(t) & ϕ′_{2}(t) & ... & ϕ′_{n}(t) \\ & . \\ & . \\ & . \\ ϕ_{1}^{(n−1)}(t) & ϕ_{2}^{(n−1)}(t) & ... & ϕ_{n}^{(n−1)}(t) \end{matrix} ]

Theorem: Suppose that ϕ_{i}, i = 1, ..., n are solutions to

y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0.

If W (ϕ_{1}, ϕ_{2}, ..., ϕ_{n})(t_{0}) ≠ 0, then there is a unique choice of constants c_{i} such that c_{1}ϕ_{1} + c_{2}ϕ_{2} + ... + c_{n}ϕ_{n} satisfies this homogeneous linear differential equation and initial conditions, y(t_{0}) = y_{0}, y′(t_{0}) = y_{1}, ..., y^{(n−1)}(t_{0}) = y_{n−1}.

Recall ϕ_{1}, ..., ϕ_{n} are linearly independent iff c_{1} = ... = c_{n} = 0 is the only solution to c_{1}ϕ_{1} + ... + c_{n}ϕ_{n} = 0.

If ϕ_{i} are functions of t, then 0 is the constant function, 0(t) = 0 for all t. Thus c_{1}ϕ_{1}(t) + ... + c_{n}ϕ_{n}(t) = 0 for all t.

Hence c_{1}ϕ_{1}^{(k)}(t) + ... + c_{n}ϕ_{n}^{(k)}(t) = 0 for all t, k if derivatives exist.

Thus ϕ_{1}, ..., ϕ_{n} are linearly independent iff for any given f, c_{1}ϕ_{1}(t) + ... + c_{n}ϕ_{n}(t) = 0 has a unique solution (that works for all t).

iff the following system of equations has a unique solution

c_{1}ϕ_{1}(t) + c_{2}ϕ_{2}(t) + ... + c_{n}ϕ_{n}(t) = 0
c_{1}ϕ′_{1}(t) + c_{2}ϕ′_{2}(t) + ... + c_{n}ϕ′_{n}(t) = 0

. . .

c_{1}ϕ_{1}^{(n−1)}(t) + c_{2}ϕ_{2}^{(n−1)}(t) + ... + c_{n}ϕ_{n}^{(n−1)}(t) = 0

iff the following system of equations has a unique solution

[ \begin{matrix}ϕ_{1}(t) & ϕ_{2}(t) & ... & ϕ_{n}(t) \\ ϕ′_{1}(t) & ϕ′_{2}(t) & ... & ϕ′_{n}(t) \\ & . \\ & . \\ & . \\ ϕ_{1}^{(n−1)}(t) & ϕ_{2}^{(n−1)}(t) & ... & ϕ_{n}^{(n−1)}(t) \end{matrix} ] [ \begin{matrix}c_{1} \\ c_{2} \\ . \\ . \\ . \\ c_{n} \end{matrix} ] = [ \begin{matrix}0 \\ 0 \\ . \\ . \\ . \\ 0 \end{matrix} ]

Note this equation has a unique solution if and only if for some t_{0}

det [ \begin{matrix}ϕ_{1}(t_{0}) & ϕ_{2}(t_{0}) & ... & ϕ_{n}(t_{0}) \\ ϕ′_{1}(t_{0}) & ϕ′_{2}(t_{0}) & ... & ϕ′_{n}(t_{0}) \\ & . \\ & . \\ & . \\ ϕ_{1}^{(n−1)}(t_{0}) & ϕ_{2}^{(n−1)}(t_{0}) & ... & ϕ_{n}^{(n−1)}(t_{0}) \end{matrix} ] ≠ 0

iff W (ϕ_{1}, ϕ_{2}, ..., ϕ_{n})(t_{0}) ≠ 0,

Example: Determine if {1+2t, 5+4t^{2}, 6−8t+8t^{2}} are linearly independent:

Method 1: Solve c_{1}(1 + 2t) + c_{2}(5 + 4t^{2}) + c_{3}(6 − 8t + 8t^{2}) = 0

Or equivalently, solve c_{1} [ \begin{matrix}1 \\ 2 \\ 0 \end{matrix} ] + c_{2} [ \begin{matrix}5 \\ 0 \\ 4 \end{matrix} ] + c_{3} [ \begin{matrix}6 \\ −8 \\ 8 \end{matrix} ] = [ \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ]

Or equivalently, solve [ \begin{matrix}1 & 5 & 6 \\ 2 & 0 & −8 \\ 0 & 4 & 8 \end{matrix} ] [ \begin{matrix}c_{1} \\ c_{2} \\ c_{3} \end{matrix} ] = [ \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ]

Method 2: Check the Wronskian

nth order LINEAR differential equation:

Theorem 4.1.1: If p_{i} : (a, b) → R, i = 1, ..., n and g : (a, b) → R are continuous and a < t_{0} < b, then there exists a unique function y = ϕ(t), ϕ : (a, b) → R that satisfies the initial value problem

y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = g(t),

y(t_{0}) = y_{0}, y′(t_{0}) = y_{1}, y^{(n−1)}(t_{0}) = y_{n−1}

Proof: We proved the case n = 1 using an integrating factor. When n > 1, see more advanced textbook.

Claim: If p_{i} are continuous on (a, b), if ϕ_{1}, ..., ϕ_{n} are linearly independent solutions to

y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0,

then {ϕ_{1}, ..., ϕ_{n}} is a basis for the solution set to this differential equation.

Theorem 4.1.2: If p_{i} are continuous on (a, b), suppose that ϕ_{i}, i = 1, ..., n are solutions to y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0. If W (ϕ_{1}, ϕ_{2}, ..., ϕ_{n})(t_{0}) ≠ 0, for some t_{0} ∈ (a, b), then any solution to this homogeneous linear differential equation can be written as

y = c_{1}ϕ_{1} + c_{2}ϕ_{2} + ... + c_{n}ϕ_{n} for some constants c_{i}.

Defn: The ϕ_{1}, ..., ϕ_{n} are called a fundamental set of solutions to y^{(n)} + p_{1}(t)y^{(n−1)} + ... + p_{n−1}(t)y′ + p_{n}(t)y = 0.

Theorem: Given any nth order homogeneous linear differential equation, there exist a set of n functions which form a fundamental set of solutions.