6 Ordinary Differential Equations

Basic Concepts of Differential Equations

• Differential Equation (System): An equation (or system of equations) containing unknown functions and their derivatives.
• Order of a Differential Equation (System): The order of the highest-order derivative of the unknown functions involved in the equation.
• Ordinary Differential Equation (ODE): A differential equation (or system) where all unknown functions are functions of a single independent variable.

General Form of an $n$-th Order ODE (Implicit Form)

$$\begin{equation*} F(x, \mathbf{y}, \mathbf{y}', \dots, \mathbf{y}^{(n)}) = 0 \end{equation*}$$

Where $\mathbf{y}(x) = (y_1(x), \dots, y_m(x))$ is a (set of) unknown function(s).

Explicit Form of a Differential Equation

$$\begin{equation*} \mathbf{y}^{(n)} = f(x, \mathbf{y}, \mathbf{y}', \dots, \mathbf{y}^{(n-1)}) \end{equation*}$$

Solution of a Differential Equation

A solution of the differential equation $F(x, \mathbf{y}, \mathbf{y}', \dots, \mathbf{y}^{(n)}) = 0$ is a function $\mathbf{y} = \mathbf{y}(x)$ that satisfies the equation for all $x$ in its domain:

$$\begin{equation*} F(x, \mathbf{y}(x), \mathbf{y}'(x), \dots, \mathbf{y}^{(n)}(x)) = 0, \quad \forall x \end{equation*}$$

Initial Value Problem (Cauchy Problem)

An Initial Value Problem (also known as a Cauchy Problem) consists of a differential equation along with a set of initial conditions at a specific point $x_0$:

$$\begin{equation*} \begin{cases} F(x, \mathbf{y}, \mathbf{y}', \dots, \mathbf{y}^{(n)}) = 0, \\ \mathbf{y}(x_0) = \mathbf{C}_0, \\ \mathbf{y}'(x_0) = \mathbf{C}_1, \\ \vdots \\ \mathbf{y}^{(n-1)}(x_0) = \mathbf{C}_{n-1} \end{cases} \end{equation*}$$

Where $\mathbf{C}_0, \mathbf{C}_1, \dots, \mathbf{C}_{n-1}$ are given constant vectors. The goal is to find a specific solution $\mathbf{y} = \mathbf{y}(x)$ that satisfies both the equation and these conditions.

For an $n$-th order differential equation $F(x, y, y', \dots, y^{(n)}) = 0$, the General Solution $y(x)$ is a family of functions that contains $n$ arbitrary constants:

Direction Fields and Integral Curves: Geometric Meaning of First-Order Differential Equations

$$\begin{equation*} \frac{dy}{dx} = f(x,y), \quad y = y(x) \text{ is a solution } \Longleftrightarrow y'(x) = f(x, y(x)). \end{equation*}$$

Slope field: At each point $(x,y)$, a slope value $f(x,y)$ is prescribed.

Direction field: At each point $(x,y)$, a line with direction

$$\begin{equation*} (1,\, f(x,y)) \end{equation*}$$

is assigned.

Integral curve: A curve $\gamma$ is called an integral curve if, at every point $(x,y)$ it passes through, the slope of its tangent line equals

$$\begin{equation*} f(x,y). \end{equation*}$$

General Definition of Direction Fields and Integral Curves

Direction field (line field): At each point $P$ in space, a straight line $l(P)$ is assigned.

Integral curve of a direction field: A curve $\gamma$ is called an integral curve if it is everywhere tangent to the direction field.

Planar Direction Field

$$\begin{equation*} P(x,y)\,dx + Q(x,y)\,dy = 0. \end{equation*}$$

Integral Curve

Let an integral curve be parametrized as

$$\begin{equation*} (x(t),\, y(t)). \end{equation*}$$

Then it satisfies

$$\begin{equation*} P\bigl(x(t),y(t)\bigr)\,x'(t) \;+\; Q\bigl(x(t),y(t)\bigr)\,y'(t) = 0, \quad \forall\, t. \end{equation*}$$

Separable Differential Equations

$$\begin{equation*} p(x)\,dx + q(y)\,dy = 0. \end{equation*}$$

Assume that $p$ and $q$ are continuous and admit antiderivatives $P$ and $Q$, respectively.

Let an integral curve be parametrized by

$$\begin{equation*} (x(t),\, y(t)). \end{equation*}$$

Then along the curve,

$$\begin{equation*} 0 = p\bigl(x(t)\bigr)\,dx(t) + q\bigl(y(t)\bigr)\,dy(t) \end{equation*}$$ $$\begin{equation*} = dP\bigl(x(t)\bigr) + dQ\bigl(y(t)\bigr) \end{equation*}$$ $$\begin{equation*} = \bigl(P(x(t)) + Q(y(t))\bigr)' \, dt. \end{equation*}$$

Therefore,

$$\begin{equation*} P(x(t)) + Q(y(t)) = C. \end{equation*}$$

Equivalently, the general solution can be written as

$$\begin{equation*} P(x) + Q(y) = C. \end{equation*}$$

Homogeneous Differential Equations

$$\begin{equation*} y' = f\!\left(\frac{y}{x}\right). \end{equation*}$$

Introduce the substitution

$$\begin{equation*} p = \frac{y}{x}, \qquad z = \ln |x|. \end{equation*}$$

Then we obtain

$$\begin{equation*} z'_p = \frac{x'_p}{x} = \frac{1}{x p'_x} = \frac{1}{y'_x - \dfrac{y}{x}} = \frac{1}{f(p) - p}. \end{equation*}$$

(This is the simplest differential equation.)

Integrating, we get

$$\begin{equation*} z = \int \frac{dp}{f(p) - p} = F(p) + C. \end{equation*}$$

Since $z = \ln |x|$, it follows that

$$\begin{equation*} x = C_1 e^{F(p)} = C_1 e^{F(y/x)}. \end{equation*}$$

Equivalently,

$$\begin{equation*} x\,e^{-F(y/x)} = C_1. \end{equation*}$$

Differential Equations of the Form

$$\begin{equation*} y' = f\!\left( \frac{a_1 x + b_1 y + c_1} {a_2 x + b_2 y + c_2} \right). \end{equation*}$$

Case 1

Assume that the linear system

$$\begin{equation*} \begin{cases} a_1 x + b_1 y + c_1 = 0, \\ a_2 x + b_2 y + c_2 = 0 \end{cases} \end{equation*}$$

has a solution $(x_0, y_0)$.

This point serves as a center of symmetry of the direction field.

Translation of Variables

Introduce the change of variables

$$\begin{equation*} \xi = x - x_0, \qquad \eta = y - y_0. \end{equation*}$$

Then the differential equation becomes

$$\begin{equation*} \eta'_\xi = f\!\left( \frac{a_1 \xi + b_1 \eta} {a_2 \xi + b_2 \eta} \right) = f\!\left( \frac{a_1 + b_1 \dfrac{\eta}{\xi}} {a_2 + b_2 \dfrac{\eta}{\xi}} \right). \end{equation*}$$

This is a homogeneous differential equation.

Case 2

Assume that the linear system

$$\begin{equation*} \begin{cases} a_1 x + b_1 y + c_1 = 0, \\ a_2 x + b_2 y + c_2 = 0 \end{cases} \end{equation*}$$

has no solution.

The differential equation can be rewritten as

$$\begin{equation*} y' = f\!\left( \alpha + \frac{\beta}{a_2 x + b_2 y + c_2} \right), \end{equation*}$$

where $\alpha$ and $\beta$ are constants.

Change of Variables

Introduce the new variable

$$\begin{equation*} \xi = a_2 x + b_2 y + c_2. \end{equation*}$$

Then

$$\begin{equation*} \xi'_x = a_2 + b_2 y'_x = a_2 + b_2 f\!\left(\alpha + \frac{\beta}{\xi}\right). \end{equation*}$$

This is a separable differential equation.

Linear Differential Equations

$$\begin{equation*} y' = a(x)\,y + f(x). \end{equation*}$$

Here $f(x)$ is the nonhomogeneous term.

• If $f(x) = 0$, the equation is called a homogeneous linear equation.
• If $f(x) \neq 0$, the equation is called a nonhomogeneous linear equation.

Homogeneous Linear Equation

The homogeneous equation

$$\begin{equation*} y' = a(x)\,y \end{equation*}$$

is a separable differential equation.

Separating variables, we obtain

$$\begin{equation*} \frac{dy}{y} = a(x)\,dx \end{equation*}$$

which implies

$$\begin{equation*} \ln |y| = \int_{x_0}^{x} a(t)\,dt + C_1. \end{equation*}$$

Equivalently,

$$\begin{equation*} |y| = e^{C_1}\, e^{\int_{x_0}^{x} a(t)\,dt}. \end{equation*}$$

Hence the general solution is

$$\begin{equation*} y(x) = C\, e^{\int_{x_0}^{x} a(t)\,dt}, \end{equation*}$$

where $C$ is an arbitrary constant. This family of solutions includes the particular solution $y = 0$.

Nonhomogeneous Linear Differential Equations

$$\begin{equation*} y' = a(x)\,y + f(x). \end{equation*}$$

Assume a solution of the form

$$\begin{equation*} y(x) = C(x)\, e^{\int_{x_0}^{x} a(t)\,dt}. \end{equation*}$$

Then

$$\begin{equation*} f(x) = y'(x) - a(x)\,y(x). \end{equation*}$$

Substituting $y(x)$ into the equation, we obtain

$$\begin{equation*} \begin{aligned} f(x) &= C'(x)\, e^{\int_{x_0}^{x} a(t)\,dt} + C(x)\, e^{\int_{x_0}^{x} a(t)\,dt} a(x) - a(x)\, C(x)\, e^{\int_{x_0}^{x} a(t)\,dt} \\ &= C'(x)\, e^{\int_{x_0}^{x} a(t)\,dt}. \end{aligned} \end{equation*}$$

Hence,

$$\begin{equation*} C'(x) = f(x)\, e^{-\int_{x_0}^{x} a(t)\,dt}. \end{equation*}$$

Integrating,

$$\begin{equation*} C(x) = C_0 + \int_{x_0}^{x} f(u)\, e^{-\int_{x_0}^{u} a(t)\,dt}\,du. \end{equation*}$$

Finally, the general solution is

$$\begin{equation*} y(x) = C_0\, e^{\int_{x_0}^{x} a(t)\,dt} + e^{\int_{x_0}^{x} a(t)\,dt} \int_{x_0}^{x} f(u)\, e^{-\int_{x_0}^{u} a(t)\,dt}\,du. \end{equation*}$$

Example

1. $$\begin{equation*} y' - y = e^{x}x^{2} - 2e^{2x}x + x\sin x + e^{-x}\cos x. \end{equation*}$$

Reduction of Higher-Order Differential Equations to First-Order Systems

An $n$-th Order Differential Equation

$$\begin{equation*} F\!\left(x, y, y', y^{(2)}, \dots, y^{(n)}\right) = 0. \end{equation*}$$

If the equation can be solved for the highest derivative, it can be written as

$$\begin{equation*} y^{(n)} = f\!\left(x, y, y', y^{(2)}, \dots, y^{(n-1)}\right). \end{equation*}$$

Define

$$\begin{equation*} u_k(x) = y^{(k)}(x), \qquad k = 0,1,\dots,n-1. \end{equation*}$$

The $n$-th order equation is equivalent to the system

$$\begin{equation*} \begin{cases} u_0' = u_1, \\ u_1' = u_2, \\ \ \vdots \\ F\!\left(x, u_0, u_1, \dots, u_{n-1}, u'_{n - 1}\right) = 0. \end{cases} \end{equation*}$$

If the highest derivative is explicitly given, we obtain

$$\begin{equation*} \begin{cases} u_0' = u_1, \\ u_1' = u_2, \\ \ \vdots \\ u'_{n - 1} = f\!\left(x, u_0, u_1, \dots, u_{n-1}\right). \end{cases} \end{equation*}$$

Linear Systems of Differential Equations with Constant Coefficients

Consider the linear system

$$\begin{equation*} \mathbf{v}' = A\,\mathbf{v}, \end{equation*}$$

where $A$ is a constant matrix.

Change of Variables

Let

$$\begin{equation*} \mathbf{v} = P\,\mathbf{u}, \end{equation*}$$

where $P$ is an invertible matrix. Then

$$\begin{equation*} \mathbf{u}' = P^{-1}\mathbf{v}' = P^{-1}A\mathbf{v} = P^{-1}AP\,\mathbf{u}. \end{equation*}$$

Hence, under the linear transformation $\mathbf{v}=P\mathbf{u}$, the system is transformed into

$$\begin{equation*} \mathbf{u}' = \left(P^{-1}AP\right)\mathbf{u}. \end{equation*}$$

Order Reduction of Higher-Order Differential Equations

Differential Equations Not Explicitly Involving $y$

Consider an $n$-th order differential equation of the form

$$\begin{equation*} F\!\left(x, y^{(k)}, y^{(k+1)}, \dots, y^{(n)}\right) = 0. \end{equation*}$$

Introduce the new unknown function

$$\begin{equation*} u(x) = y^{(k)}(x). \end{equation*}$$

Then the original equation is transformed into the system

$$\begin{equation*} \begin{cases} F\!\left(x, u, u', \dots, u^{(n-k)}\right) = 0, \\ y^{(k)} = u(x). \end{cases} \end{equation*}$$

The order of the differential equation is thus reduced by $k$.

Differential Equations Not Explicitly Involving $x$

Consider an $n$-th order differential equation of the form

$$\begin{equation*} F\!\left(y, y', \dots, y^{(n)}\right) = 0. \end{equation*}$$

Introduce the substitution

$$\begin{equation*} u(y) = y'(x). \end{equation*}$$

For any differentiable function $w = w(y) = w(y(x))$, by the chain rule we have

$$\begin{equation*} \frac{dw}{dx} = \frac{dw}{dy}\frac{dy}{dx} = u\,\frac{dw}{dy}, \end{equation*}$$

that is,

$$\begin{equation*} \frac{d}{dx} = u\,\frac{d}{dy}. \end{equation*}$$

Using this relation, the original equation is transformed into

$$\begin{equation*} \begin{cases} F\!\left( y,\, u,\, u\,\dfrac{du}{dy},\, \dots,\, \left(u\dfrac{d}{dy}\right)^{n-1}u \right) = 0, \\ \dfrac{dy}{dx} = u(y). \end{cases} \end{equation*}$$

Factorizable Linear Differential Equations

Consider the second-order linear differential equation

$$\begin{equation*} Ly = y'' - 3y' + 2y = x^2. \end{equation*}$$

Introduce the differential operator

$$\begin{equation*} D = \frac{d}{dx}, \end{equation*}$$

then

$$\begin{equation*} L = D^2 - 3D + 2. \end{equation*}$$

Factor the characteristic polynomial:

$$\begin{equation*} \lambda^2 - 3\lambda + 2 = (\lambda - 2)(\lambda - 1). \end{equation*}$$

Accordingly, the differential operator can be factorized as

$$\begin{equation*} L = \left(\frac{d}{dx} - 2\right) \left(\frac{d}{dx} - 1\right). \end{equation*}$$

Thus the equation becomes

$$\begin{equation*} \left(\frac{d}{dx} - 2\right) \left(\frac{d}{dx} - 1\right)y = x^2. \end{equation*}$$

Define

$$\begin{equation*} u = \left(\frac{d}{dx} - 1\right)y = y' - y. \end{equation*}$$

Then the original equation is equivalent to the system

$$\begin{equation*} \begin{cases} u' - 2u = x^2, \\ y' - y = u. \end{cases} \end{equation*}$$

Factorizable Differential Equations (Euler Type)

Consider the Euler-type differential equation

$$\begin{equation*} Ly = x^2 y'' - 2x y' + 2y = f(x). \end{equation*}$$

Introduce the Euler differential operator

$$\begin{equation*} D_x = x\frac{d}{dx}. \end{equation*}$$

Then the operator $L$ can be written as

$$\begin{equation*} L = (D_x - \alpha)(D_x - \beta). \end{equation*}$$

Thus,

$$\begin{equation*} Ly = \left(x\frac{d}{dx} - \alpha\right) \left(x\frac{d}{dx} - \beta\right)y = f(x). \end{equation*}$$

Expanding the operator and comparing coefficients yields

$$\begin{equation*} \begin{cases} x - \alpha x - \beta x = -2x, \\ \alpha\beta = 2. \end{cases} \end{equation*}$$

Solving this system gives

$$\begin{equation*} \alpha = 2, \qquad \beta = 1. \end{equation*}$$

Define

$$\begin{equation*} u = \left(x\frac{d}{dx} - \beta\right)y = x y' - y. \end{equation*}$$

Then the original equation is equivalent to the system

$$\begin{equation*} \begin{cases} x u' - 2u = f(x), \\ x y' - y = u. \end{cases} \end{equation*}$$

High-Order Linear Differential Equation and First-Order Linear System

Consider the (n)-th order linear differential equation

$$\begin{equation*} y^{(n)} + a_{n-1}(x)y^{(n-1)} + \cdots + a_1(x)y' + a_0(x)y = f(x). \end{equation*}$$

Introduce the state variables

$$\begin{equation*} \begin{aligned} u_0 &= y, \\ u_1 &= y', \\ &\;\;\vdots \\ u_{n-2} &= y^{(n-2)}, \\ u_{n-1} &= y^{(n-1)}. \end{aligned} \end{equation*}$$

Then the equation can be rewritten as the first-order linear system

$$\begin{equation*} \begin{pmatrix} u_0 \\ u_1 \\ \vdots \\ u_{n-2} \\ u_{n-1} \end{pmatrix}' = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_0(x) & -a_1(x) & -a_2(x) & \cdots & -a_{n-1}(x) \end{pmatrix} \begin{pmatrix} u_0 \\ u_1 \\ \vdots \\ u_{n-2} \\ u_{n-1} \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ f(x) \end{pmatrix}. \end{equation*}$$

That is, the high-order linear differential equation is equivalent to a first-order linear differential system.

Superposition Principle for Linear Differential Systems

Homogeneous vs. Nonhomogeneous Systems

Let $A(x)$ be a matrix-valued function.

• Homogeneous linear system

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y} \end{equation*}$$

• Nonhomogeneous linear system

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y} + \mathbf{f}(x) \end{equation*}$$

Here $\mathbf{y}(x)$ is a vector-valued function and $\mathbf{f}(x)$ is the nonhomogeneous (forcing) term.

Method of Variation of Constants (for Linear Systems)

Homogeneous Linear System

Consider the homogeneous linear system

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y}, \end{equation*}$$

where $A(x)$ is an $n \times n$ matrix-valued function.

Let $\mathbf{y}_1(x), \ldots, \mathbf{y}_n(x)$ be $n$ linearly independent solutions of this system. Define the fundamental matrix

$$\begin{equation*} U(x) = \bigl(\mathbf{y}_1(x), \ldots, \mathbf{y}_n(x)\bigr). \end{equation*}$$

Then $U(x)$ is invertible and satisfies

$$\begin{equation*} U'(x) = A(x)U(x). \end{equation*}$$

Nonhomogeneous Linear System

Now consider the nonhomogeneous system

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y} + \mathbf{f}(x). \end{equation*}$$

Assume a solution of the form

$$\begin{equation*} \mathbf{y}(x) = U(x)\mathbf{C}(x), \end{equation*}$$

where $\mathbf{C}(x)$ is an unknown vector-valued function.

Differentiate $\mathbf{y}(x)$:

$$\begin{equation*} \begin{aligned} \mathbf{y}'(x) &= U'(x)\mathbf{C}(x) + U(x)\mathbf{C}'(x) \\ &= A(x)U(x)\mathbf{C}(x) + U(x)\mathbf{C}'(x) \\ &= A(x)\mathbf{y}(x) + U(x)\mathbf{C}'(x). \end{aligned} \end{equation*}$$

Comparing with

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y} + \mathbf{f}(x), \end{equation*}$$

we obtain

$$\begin{equation*} U(x)\mathbf{C}'(x) = \mathbf{f}(x). \end{equation*}$$

Since $U(x)$ is invertible,

$$\begin{equation*} \mathbf{C}'(x) = U(x)^{-1}\mathbf{f}(x). \end{equation*}$$

Integrating,

$$\begin{equation*} \mathbf{C}(x) = \mathbf{C}_0 + \int_{x_0}^x U(t)^{-1}\mathbf{f}(t)\,dt. \end{equation*}$$

Thus the general solution of the nonhomogeneous system is

$$\begin{equation*} \mathbf{y}(x) = U(x)\mathbf{C}_0 + U(x)\int_{x_0}^x U(t)^{-1}\mathbf{f}(t)\,dt. \end{equation*}$$

Interpretation

• $U(x)\mathbf{C}_0$ is the general solution of the homogeneous system.
• The integral term provides one particular solution of the nonhomogeneous system.
• This method generalizes the variation of constants technique from scalar equations to linear systems.

Let $\mathbf{y}_1,\mathbf{y}_2$ be two solutions of the linear differential equation

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y} + \mathbf{f}(x), \end{equation*}$$

satisfying

$$\begin{equation*} \mathbf{y}_1(x_0) = \mathbf{y}_2(x_0). \end{equation*}$$

Then

$$\begin{equation*} \mathbf{y}_1(x) = \mathbf{y}_2(x), \qquad \forall x. \end{equation*}$$

Liouville’s Theorem. Let $\mathbf{y}_1,\dots,\mathbf{y}_n$ be $n$ solutions of the homogeneous linear system

$$\begin{equation*} \mathbf{y}' = A(x)\mathbf{y}. \end{equation*}$$

Define the matrix

$$\begin{equation*} U(x) = \bigl(\mathbf{y}_1(x),\dots,\mathbf{y}_n(x)\bigr). \end{equation*}$$

Then $U(x)$ satisfies

$$\begin{equation*} \bigl(\det U(x)\bigr)' = \operatorname{tr} A(x)\,\det U(x). \end{equation*}$$

Corollary.

For any $x$, the functions $\mathbf{y}_1(x),\dots,\mathbf{y}_n(x)$ are linearly independent if and only if there exists a point $x_0$ such that $\mathbf{y}_1(x_0),\dots,\mathbf{y}_n(x_0)$ are linearly independent.