MathIsimple
ODE-04
Higher-Order Methods

Higher-Order Linear Equations

Learn powerful techniques for reducing the order of differential equations, understanding first integrals, analyzing autonomous systems, and converting higher-order equations to first-order systems.

Learning Objectives

  • Apply reduction of order for missing variables
  • Identify and use first integrals
  • Solve autonomous (stationary) equations
  • Perform phase plane analysis
  • Convert higher-order ODEs to systems
  • Understand the structure of solution spaces
  • Solve physical systems using energy methods
  • Apply these techniques to tracking problems

1. Reduction of Order

For higher-order equations, reducing the order can significantly simplify the problem. The key insight is that when certain variables are missing from the equation, we can introduce substitutions that lower the order.

Definition 4.1: Order Reduction for Missing Variables

Consider the nn-th order equation:

F(t,x,x,,x(n))=0F(t, x, x', \ldots, x^{(n)}) = 0

Case 1: If the equation has the form F(t,x(k),,x(n))=0F(t, x^{(k)}, \ldots, x^{(n)}) = 0 (missing x,x,,x,(k1)x, x', \ldots, x^,{(k-1)}), substitute y=x,(k)y = x^,{(k)} to get an (nk)(n-k)-th order equation.

Case 2: If the equation has the form F(x,x,,x(n))=0F(x, x', \ldots, x^{(n)}) = 0 (missing tt), it is called autonomous. Substitute y=xy = x' and use x=ydydxx'' = y\frac{dy}{dx}.

Theorem 4.1: Reduction for Missing Independent Variable

If tt does not appear explicitly in F(x,x,,x(n))=0F(x, x', \ldots, x^{(n)}) = 0, then setting y=xy = x'and expressing higher derivatives as:

x=ydydx,x=yddx(ydydx),x'' = y\frac{dy}{dx}, \quad x''' = y\frac{d}{dx}\left(y\frac{dy}{dx}\right), \quad \ldots

reduces the nn-th order equation to an (n1)(n-1)-th order equation in yy as a function of xx.

Proof of Theorem 4.1:

Since y=x=dx/dty = x' = dx/dt, by the chain rule:

x=dydt=dydxdxdt=ydydxx'' = \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = y\frac{dy}{dx}

For the third derivative:

x=ddt(ydydx)=ddx(ydydx)dxdt=yddx(ydydx)x''' = \frac{d}{dt}\left(y\frac{dy}{dx}\right) = \frac{d}{dx}\left(y\frac{dy}{dx}\right) \cdot \frac{dx}{dt} = y\frac{d}{dx}\left(y\frac{dy}{dx}\right)

By induction, each x,(k)x^,{(k)} can be expressed in terms of yy and its derivatives with respect to xx. ∎

Algorithm 4.1: Reduction of Order Method

Input: Higher-order ODE F(t,x,x,,x(n))=0F(t, x, x', \ldots, x^{(n)}) = 0

Output: General solution

1. Check which variables are missing from the equation

2. If x,x,,x,(k1)x, x', \ldots, x^,{(k-1)} are missing:

2.1. Set y=x,(k)y = x^,{(k)}

2.2. Solve the reduced equation for y(t)y(t)

2.3. Integrate kk times to find x(t)x(t)

3. If tt is missing (autonomous):

3.1. Set y=xy = x', express x=ydy/dxx'' = y \cdot dy/dx

3.2. Solve for y(x)y(x)

3.3. Integrate dx/dt=y(x)dx/dt = y(x) to find x(t)x(t)

4. return General solution with appropriate constants

Example 4.1: Equation Missing Lower Derivatives

Problem: Solve t2x+tx=1t^2 x'' + tx' = 1.

Solution:

The equation is missing xx. Set y=xy = x', so y=xy' = x'':

t2y+ty=1t^2 y' + ty = 1

This is a first-order linear equation. Dividing by t2t^2:

y+1ty=1t2y' + \frac{1}{t}y = \frac{1}{t^2}

The integrating factor is μ=e(1/t)dt=t\mu = e^{\int (1/t)dt} = t. Multiplying:

(ty)=1t    ty=lnt+C1    y=lnt+C1t(ty)' = \frac{1}{t} \implies ty = \ln|t| + C_1 \implies y = \frac{\ln|t| + C_1}{t}

Integrating to find xx:

x=lnt+C1tdt=12(lnt)2+C1lnt+C2x = \int \frac{\ln|t| + C_1}{t}\,dt = \frac{1}{2}(\ln|t|)^2 + C_1\ln|t| + C_2

2. Autonomous Equations

Definition 4.2: Autonomous Equation

A differential equation is called autonomous (or stationary) if the independent variable tt does not appear explicitly:

F(x,x,x,,x(n))=0F(x, x', x'', \ldots, x^{(n)}) = 0

The term "autonomous" comes from the fact that the system's behavior is independent of the absolute time—only the state matters.

Example 4.2: Physical System: Newton's Second Law

Problem: Solve mx=f(x)mx'' = f(x) where mm is mass and f(x)f(x) is a position-dependent force.

Solution:

This is autonomous (no explicit tt). Set p=xp = x' (momentum per unit mass):

x=dpdt=dpdxdxdt=pdpdxx'' = \frac{dp}{dt} = \frac{dp}{dx}\frac{dx}{dt} = p\frac{dp}{dx}

The equation becomes:

mpdpdx=f(x)    mpdp=f(x)dxmp\frac{dp}{dx} = f(x) \implies m\,p\,dp = f(x)\,dx

Integrating both sides:

12mp2=f(x)dx+C\frac{1}{2}mp^2 = \int f(x)\,dx + C

This is the energy integral: 12m(x)2f(x)dx=C\frac{1}{2}m(x')^2 - \int f(x)dx = C, expressing conservation of mechanical energy (kinetic + potential = constant).

Remark:

The energy integral is a first integral of the equation. For conservative mechanical systems, multiplying the equation of motion by xx' and integrating always yields the energy conservation law.

3. First Integrals

Definition 4.3: First Integral

For the system dxdt=f(t,x)\frac{dx}{dt} = f(t, x) where xRnx \in \mathbb{R}^n, a functionΦ(t,x)C1(D)\Phi(t, x) \in C^1(D) is called a first integral if:

  1. Φ\Phi is not constant on DD
  2. Φ\Phi is constant along every solution curve in DD

Equivalently, dΦdt=Φt+xΦf=0\frac{d\Phi}{dt} = \frac{\partial \Phi}{\partial t} + \nabla_x \Phi \cdot f = 0along solutions.

Definition 4.4: Independent First Integrals

First integrals Φ1,,Φn\Phi_1, \ldots, \Phi_n are called independent if their Jacobian with respect to the state variables is non-zero:

(Φ1,,Φn)(x1,,xn)=Φ1x1Φ1xnΦnx1Φnxn0\frac{\partial(\Phi_1, \ldots, \Phi_n)}{\partial(x_1, \ldots, x_n)} = \begin{vmatrix} \frac{\partial \Phi_1}{\partial x_1} & \cdots & \frac{\partial \Phi_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial \Phi_n}{\partial x_1} & \cdots & \frac{\partial \Phi_n}{\partial x_n} \end{vmatrix} \neq 0

Theorem 4.2: Existence and Uniqueness of First Integrals

If fC1(D)f \in C^1(D) where DRn+1D \subset \mathbb{R}^{n+1}, then in a neighborhood of any regular point, the system dxdt=f(t,x)\frac{dx}{dt} = f(t, x) has exactly nn independent first integrals.

Proof of Theorem 4.2:

Existence: Consider the initial value problem with x(t0)=C=(C1,,Cn)Tx(t_0) = C = (C_1, \ldots, C_n)^T. Let Φ(t,C)\Phi(t, C) denote the solution. By the implicit function theorem, we can locally solveCj=Ψj(t,x)C_j = \Psi_j(t, x) for each jj. These Ψj\Psi_j are first integrals.

Uniqueness (at most nn): Taking the total derivative of any first integral:

dΨjdt=Ψjt+k=1nΨjxkfk=0\frac{d\Psi_j}{dt} = \frac{\partial \Psi_j}{\partial t} + \sum_{k=1}^{n} \frac{\partial \Psi_j}{\partial x_k} f_k = 0

For n+1n+1 first integrals, the vector (1,f1,,fn)(1, f_1, \ldots, f_n) would be a non-trivial solution to a homogeneous system, implying the Jacobian determinant is zero—contradicting independence. ∎

Theorem 4.3: Finding General Solution from First Integrals

If Φ1(t,x)=C1,,Φn(t,x)=Cn\Phi_1(t, x) = C_1, \ldots, \Phi_n(t, x) = C_n are nn independent first integrals of dxdt=f(t,x)\frac{dx}{dt} = f(t, x), then the general solution can be obtained by solving this system of algebraic equations for xx in terms of tt and the constants C1,,CnC_1, \ldots, C_n.

Example 4.3: Finding First Integrals

Problem: Find first integrals for the system x=yx' = -y, y=xy' = x.

Solution:

This system describes rotation. Multiply the first equation by xx and the second by yy:

xdxdt+ydydt=xy+xy=0x\frac{dx}{dt} + y\frac{dy}{dt} = -xy + xy = 0

This gives:

12ddt(x2+y2)=0    x2+y2=C1\frac{1}{2}\frac{d}{dt}(x^2 + y^2) = 0 \implies x^2 + y^2 = C_1

For the second first integral, from y=xy' = x we have x=yx = y'. Substituting:

dydt=x    dyC1y2=dt\frac{dy}{dt} = x \implies \frac{dy}{\sqrt{C_1 - y^2}} = dt

Integrating: arcsin(y/C1)=t+C2\arcsin(y/\sqrt{C_1}) = t + C_2, giving the second first integral.

4. Converting Higher-Order Equations to Systems

Theorem 4.4: Conversion to First-Order System

Any explicit nn-th order ODE x(n)=f(t,x,x,,x(n1))x^{(n)} = f(t, x, x', \ldots, x^{(n-1)}) can be converted to a first-order system of dimension nn by setting:

x1=x,x2=x,,xn=x(n1)x_1 = x, \quad x_2 = x', \quad \ldots, \quad x_n = x^{(n-1)}

The resulting system is:

{x1=x2x2=x3xn1=xnxn=f(t,x1,x2,,xn)\begin{cases} x_1' = x_2 \\ x_2' = x_3 \\ \vdots \\ x_{n-1}' = x_n \\ x_n' = f(t, x_1, x_2, \ldots, x_n) \end{cases}

Example 4.4: Converting a Second-Order Equation

Problem: Convert x+3x+2x=sintx'' + 3x' + 2x = \sin t to a first-order system.

Solution:

Set x1=xx_1 = x and x2=xx_2 = x'. Then:

{x1=x2x2=2x13x2+sint\begin{cases} x_1' = x_2 \\ x_2' = -2x_1 - 3x_2 + \sin t \end{cases}

In matrix form: x=Ax+b(t)\mathbf{x}' = A\mathbf{x} + \mathbf{b}(t) where:

A=(0123),b(t)=(0sint)A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, \quad \mathbf{b}(t) = \begin{pmatrix} 0 \\ \sin t \end{pmatrix}
Remark:

Converting to a first-order system is fundamental because: (1) it unifies the theory—all results for first-order systems apply, (2) numerical methods are typically designed for first-order systems, and (3) the geometric interpretation (phase space) becomes clearer.

5. Applications: Missile Tracking Problem

Example 4.5: Missile Tracking Problem

Problem: A missile at the origin tracks a ship at (0,H)(0, H) moving east at speed vev_e. The missile travels at speed vwv_w and always points toward the ship. Find the trajectory and interception point.

Solution:

Let P(x(t),y(t))P(x(t), y(t)) be the missile position. The velocity magnitude condition:

(dxdt)2+(dydt)2=vw2\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = v_w^2

The pointing condition (tangent toward ship at (vet,H)(v_e t, H)):

dydx=Hyvetx\frac{dy}{dx} = \frac{H - y}{v_e t - x}

Differentiating the pointing condition with respect to tt and using λ=ve/vw\lambda = v_e/v_w, we get:

d2xdy2Hy1+(dx/dy)2=λ\frac{d^2x}{dy^2} \cdot \frac{H-y}{\sqrt{1 + (dx/dy)^2}} = \lambda

Setting p=dx/dyp = dx/dy, this becomes a separable first-order equation. After solving with initial conditions x(0)=0,x(0)=0x(0) = 0, x'(0) = 0:

x=12((Hy)1+λHλ(1+λ)Hλ(Hy)1λ1λ)+λH1λ2x = \frac{1}{2}\left(\frac{(H-y)^{1+\lambda}}{H^\lambda(1+\lambda)} - \frac{H^\lambda(H-y)^{1-\lambda}}{1-\lambda}\right) + \frac{\lambda H}{1-\lambda^2}

When y=Hy = H, the missile intercepts at x=L=λH1λ2x = L = \frac{\lambda H}{1-\lambda^2}, at time T=L/veT = L/v_e.

Practice Quiz

Higher-Order Equations Quiz
10
Questions
0
Correct
0%
Accuracy
1
For the equation F(t,x(k),,x(n))=0F(t, x^{(k)}, \ldots, x^{(n)}) = 0 where k1k \geq 1, what substitution reduces the order?
Easy
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2
An autonomous equation is one where:
Easy
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3
What is a first integral of a differential equation?
Easy
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4
For the autonomous equation F(x,x,x)=0F(x, x', x'') = 0, what substitution is commonly used?
Medium
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5
How many independent first integrals does an nn-th order ODE have in a neighborhood of a regular point?
Medium
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6
The equation mx=f(x)mx'' = f(x) (where mm is mass and ff is force) can be integrated once to get:
Medium
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7
A second-order ODE x=f(t,x,x)x'' = f(t, x, x') can be converted to a first-order system of dimension:
Medium
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8
In the missile tracking problem, if the missile always points toward the target, the resulting equation is:
Medium
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9
For the nn-th order equation x(n)=f(t,x,,x(n1))x^{(n)} = f(t, x, \ldots, x^{(n-1)}), setting xk=x(k1)x_k = x^{(k-1)} for k=1,,nk = 1, \ldots, n gives a system where:
Hard
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10
If Φ1,,Φn\Phi_1, \ldots, \Phi_n are nn independent first integrals of the system dxdt=f(t,x)\frac{dx}{dt} = f(t,x), then the Jacobian (Φ1,,Φn)(x1,,xn)\frac{\partial(\Phi_1, \ldots, \Phi_n)}{\partial(x_1, \ldots, x_n)} is:
Hard
Not attempted

Frequently Asked Questions

When should I use reduction of order?

Use reduction of order when the equation is missing certain variables: (1) If x,x,,x,(k1)x, x', \ldots, x^,{(k-1)}are missing, substitute y=x,(k)y = x^,{(k)}. (2) If tt is missing (autonomous equation), substitutey=xy = x' and express xx'' as y(dy/dx)y(dy/dx). This reduces the order by one.

What is the physical meaning of a first integral?

A first integral represents a conserved quantity. For mechanical systems, the most common first integral is energy: E=12m(x)2+V(x)=constantE = \frac{1}{2}m(x')^2 + V(x) = \text{constant}. Other examples include angular momentum in central force problems and the Jacobi integral in rotating systems.

Why convert higher-order equations to systems?

Converting to first-order systems: (1) Provides a unified theoretical framework, (2) Makes numerical solution straightforward (most algorithms are for first-order systems), (3) Enables phase plane analysis and geometric interpretation, (4) Connects to linear algebra methods when the system is linear.

How do I find first integrals in general?

There's no universal method, but common approaches include: (1) Identify symmetries (Noether's theorem), (2) Multiply the equation by appropriate factors and integrate, (3) For Hamiltonian systems, use the Hamiltonian itself, (4) For systems with known structure (e.g., rotation), exploit the geometry.

What is the relationship between order and dimension?

An nn-th order ODE is equivalent to a first-order system of dimension nn. The "state" of the system requires nn initial conditions: x(t0),x(t0),,x,(n1),(t0)x(t_0), x'(t_0), \ldots, x^,{(n-1)},(t_0). These become the initial values of the nn-dimensional state vector.

Can all higher-order equations be reduced?

Not all equations admit useful reductions. Reduction of order works when specific variables are missing. For general equations F(t,x,x,,x(n))=0F(t, x, x', \ldots, x^{(n)}) = 0, conversion to a first-order system is always possible, but this doesn't reduce complexity—it just changes the form.