Classification of Differential Equations

Before you can choose a solution method, you have to know what kind of equation you are looking at. Four labels do almost all of the work: the order, the degree, whether the equation is linear or nonlinear, and whether it is homogeneous or non-homogeneous. This section defines each label and then explains what it means for a function or a relation to be a solution.

Quick Reference

Classification	Definition	Example
Order	The order of the highest derivative in the equation	$5 y^{(4)} + 4 y = 0$ has order $4$
Degree	The exponent of the highest ordered derivative, after clearing radicals and fractions	(y'')^3 + y = 0 has degree $3$
Linear	The unknown and all its derivatives appear only to the first power, with no products among them	y'' + \sin(t)\, y' = \cos t
Nonlinear	Any product, power, or nonlinear function of $y$ or its derivatives	y'' + \frac{g}{L}\sin y = 0
Homogeneous	The right-hand side $f (x)$ is identically zero	y'' + y = 0
Non-homogeneous	The right-hand side $f (x)$ is not identically zero	y'' + y = e^x
Explicit solution	A function $y = ϕ (x)$ satisfying the equation on an interval	$y = \sin x$ solves y'' + y = 0
Implicit solution	A relation $Φ (x, y) = 0$ that defines such a function	$3 x^{2} + 6 x y + 6 x - 2 y^{3} + 18 y - 4 = 0$
General solution	A solution containing $n$ arbitrary constants for an $n$ th order equation	$y = c_{1} \sin x + c_{2} \cos x$
Particular solution	Obtained from the general solution by fixing the constants	$y = 2 \sin x - \cos x$
Singular solution	A solution not obtainable from the general solution for any constants	Arises only in nonlinear equations

Order of a Differential Equation

The order of a differential equation is the order of the highest derivative, or highest partial derivative, occurring in the equation.

For example,

5 y^{(4)} + 4 y = 0

is an ordinary differential equation of order $4$ , and

\frac{\partial u}{\partial t} = ρ \frac{\partial^{2} u}{\partial x^{2}}

is a partial differential equation of order $2$ .

The General $n$ th Order Form

An $n$ th order ODE can be expressed as

F\left(x, y, y', y'', \cdots, y^{(n)}\right) = 0 \tag{i}

where $F$ is some function of $n + 2$ variables: $x$ , $y$ , y', y'', up through $y^{(n)}$ . Count them: one for $x$ , one for $y$ itself, and $n$ more for the derivatives of orders $1$ through $n$ , giving $n + 2$ slots in total.

If we can solve this relation for the highest derivative, we can rewrite it in the form

y^{(n)} = G\left(x, y, y', \cdots, y^{(n-1)}\right) \tag{ii}

Form (ii) is called the normal form or solved form, and most solution techniques assume you can reach it. Solving for the highest derivative is not always possible, which is why form (i) is stated first.

A second-order PDE for $u (x, t)$ can be expressed in the same spirit as

H (x, t, u, \frac{\partial u}{\partial x}, \frac{\partial u}{\partial t}, \frac{\partial^{2} u}{\partial x^{2}}, \frac{\partial^{2} u}{\partial x \partial t}, \frac{\partial^{2} u}{\partial t^{2}}) = 0

Find the order of each equation.

$\frac{d y}{d x} + y^{5} = x$
${(\frac{d^{2} y}{d x^{2}})}^{7} + \frac{d^{3} y}{d x^{3}} = 0$
\displaystyle{y'' + \left(y'''\right)^2 + y^{10} = e^x}

Solution

Order $1$ . The only derivative is the first. The fifth power on $y$ is irrelevant to the order, since order counts derivatives, not powers of $y$ .
Order $3$ . The third derivative is present, even though the second derivative carries the large exponent. Order looks at which derivative is highest, not at exponents.
Order $3$ , for the same reason.

Degree of a Differential Equation

The degree of a differential equation is the exponent of the highest ordered derivative in it. To find the degree, we must first rationalize the equation and clear fractions.

The rationalizing step matters. Degree is only defined once the equation has been written as a polynomial in the derivatives, so radicals and denominators containing derivatives must be cleared first.

Find the order and degree of

\frac{\frac{d y}{d x}}{{[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2}} = 1 .

Solution

The highest derivative is $d y / d x$ , so the order is 1. To find the degree, clear the fractional power. Write $p = d y / d x$ and cross-multiply:

p = {(1 + p^{2})}^{3 / 2}

Squaring both sides removes the half-power:

p^{2} = {(1 + p^{2})}^{3}

The right-hand side expands to a polynomial of degree $6$ in $p$ , and it is the highest power that decides the degree. So the degree is 6.

Note that the original equation gave no hint of degree $6$ ; the rationalizing step is what revealed it.

Find the order and degree of ${(\frac{d^{2} y}{d x^{2}})}^{3} + 4 {(\frac{d y}{d x})}^{7} + y = 0$ .

Solution

The highest derivative is $d^{2} y / d x^{2}$ , so the order is $2$ . The exponent attached to that highest derivative is $3$ , so the degree is $3$ . The exponent $7$ on the first derivative plays no role, because degree looks only at the highest ordered derivative.

Order and degree answer different questions. Order asks which derivative is highest. Degree asks what power that particular derivative is raised to, once the equation is polynomial in the derivatives.

Linear vs. Nonlinear, Homogeneous vs. Non-homogeneous

The ordinary differential equation F(x, y, y', \dots, y^{(n)}) = 0 is said to be linear if the dependent variable and its derivatives appear only to the first degree. In other words, a linear ordinary differential equation of order $n$ is an equation of the form

a_n(x)\frac{d^n y}{dx^n} + a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}} + \cdots + a_1(x)\frac{dy}{dx} + a_0(x)y(x) = f(x) \tag{iii}

where the coefficients $a_{n} (x), \dots, a_{0} (x)$ and $f (x)$ are given functions of the independent variable $x$ .

Notice that if F(x, y, y', \dots, y^{(n)}) = 0 is a linear differential equation, then $F$ is a linear function of the dependent variable and its derivatives, that is, of $y$ , y', through $y^{(n)}$ . However, $F$ does not have to be a linear function of $x$ for the differential equation to be linear. The coefficients $a_{k} (x)$ and the right-hand side $f (x)$ may be as complicated as you like: $\sin x$ , $e^{x^{2}}$ , or $\ln x$ are all acceptable. Linearity is a statement about $y$ and its derivatives only.

The distinguishing characteristics of a linear ODE are:

$y, \frac{d y}{d x}, \dots, \frac{d^{n} y}{d x^{n}}$ all have exponent one.
There is no product of $y$ with any of its derivatives, no product of two derivatives, and no nonlinear function such as $\sin y$ , $e^{y}$ , or $\sqrt{y}$ applied to $y$ or a derivative.

Some examples on linear and nonlinear differential equations.

Solution

The equation $\frac{d^{2} y}{d t^{2}} - μ (1 - y^{2}) \frac{d y}{d t} + ν y = 0$ where $μ$ and $ν$ are constants is nonlinear, because it involves the product of $y^{2}$ , a nonlinear term, with $\frac{d y}{d t}$ . This is the Van der Pol equation, a standard model of a self-sustaining oscillator.
The equation $\frac{d^{2} y}{d t^{2}} + \frac{g}{L} \sin (y) = 0$ where $g$ and $L$ are constants is nonlinear because of the presence of $\sin (y)$ . This is the pendulum equation. Replacing $\sin y$ by $y$ for small angles is exactly the step that linearizes it.
The equation $\frac{d^{2} y}{d t^{2}} + \sin t \frac{d y}{d t} = \cos t$ is linear, since it has the form of Equation (iii): the coefficient $\sin t$ and the right-hand side $\cos t$ are functions of the independent variable only, and $y$ and y' each appear to the first power.

Compare the second and third examples carefully. Both contain a sine, but $\sin (y)$ makes an equation nonlinear while $\sin (t)$ leaves it linear. What is inside the sine is what matters.

Homogeneous and Non-homogeneous

If $f (x) \equiv 0$ , we say Equation (iii) is homogeneous; otherwise it is non-homogeneous.

The symbol $\equiv$ means "is identically zero", that is, zero for every $x$ in the interval, not merely at isolated points. So y'' + y = 0 is homogeneous, while y'' + y = e^x is non-homogeneous even though $e^{x}$ is never zero anywhere.

Be warned that the word "homogeneous" carries a second, unrelated meaning for first-order equations, where y' = f(y/x) is also called homogeneous. Context tells you which sense is intended.

Linearity for Partial Differential Equations

A partial differential equation is said to be linear if no power or product of the unknown function and its partial derivatives is present after rationalizing the equation and clearing fractions. For example, a linear second-order partial differential equation for $u (x, t)$ has the form

a_{1} (x, t) \frac{\partial^{2} u}{\partial x^{2}} + a_{2} (x, t) \frac{\partial^{2} u}{\partial x \partial t} + a_{3} (x, t) \frac{\partial^{2} u}{\partial t^{2}} + a_{4} (x, t) \frac{\partial u}{\partial x} + a_{5} (x, t) \frac{\partial u}{\partial t} + a_{6} (x, t) u = f (x, t)

where $a_{1} (x, t), \dots, a_{6} (x, t)$ and $f (x, t)$ are given functions.

In general, a linear partial differential equation for $u (x, t)$ has the form

\sum_{n=0}^{N} \sum_{m=0}^{M} a_{mn}(x,t) \frac{\partial^{m+n} u}{\partial x^n \, \partial t^m} = f(x,t) \tag{iv}

where $a_{m n} (x, t)$ and $f (x, t)$ are given functions of $x$ and $t$ , and $M$ and $N$ are fixed positive integers. We define $\frac{\partial^{0} u}{\partial x^{0} \partial t^{0}} = u$ , so that the $m = n = 0$ term of the sum is simply $a_{00} (x, t) u$ .

Equation (iv) is homogeneous if $f (x, t) \equiv 0$ .

Explicit and Implicit Solutions

Consider a general differential equation of order $n$ described by

F(x, y, y', \dots, y^{(n)}) = 0 . \tag{v}

A function $ϕ (x)$ is a solution of Equation (v) on an interval $I$ if substituting $ϕ (x)$ into the equation, that is, replacing $y$ by $ϕ (x)$ , y' by \phi'(x), through $y^{(n)}$ by $ϕ^{(n)} (x)$ , reduces the equation to an identity at every point of $I$ :

F\left(x, \phi(x), \phi'(x), \dots, \phi^{(n)}(x)\right) = 0 .

Such a function is said to satisfy the equation on $I$ , and the graph of $ϕ$ is called a solution curve.

The phrase "at every point of $I$ " is the whole content of the definition. A function that happens to satisfy the equation at a few scattered values of $x$ is not a solution.

For example, $y = \sin x$ is a solution of

y'' + y = 0 \tag{vi}

because

(\sin x)'' + \sin x = -\sin x + \sin x = 0 \qquad (-\infty < x < \infty) .

Similarly, $y = \cos x$ is also a solution of this differential equation. In fact, any function of the form

y = c_1 \sin x + c_2 \cos x \tag{vii}

where $c_{1}$ and $c_{2}$ are arbitrary constants is also a solution. This means the equation y'' + y = 0 has infinitely many solutions. However, not all differential equations have infinitely many solutions. A given differential equation may have a unique solution, several solutions, or no solution at all.

Implicit Solutions

Sometimes the relationship between $x$ and $y$ cannot be untangled into the form $y = ϕ (x)$ . The relation still counts as a solution.

A relation $Φ (x, y) = 0$ is called an implicit solution of the differential equation (v) on an interval $I$ if

there exists a function $y = ϕ (x)$ such that $Φ [x, ϕ (x)] = 0$ for every $x$ in $I$ , and
$ϕ (x)$ satisfies Equation (v), that is, F\left(x, \phi(x), \phi'(x), \cdots, \phi^{(n)}(x)\right) = 0 \quad \text{for every } x \in I .

For example, consider the differential equation

\frac{d y}{d x} = \frac{x + y + 1}{y^{2} - x - 3} .

We can show that

3 x^{2} + 6 x y + 6 x - 2 y^{3} + 18 y - 4 = 0

is an implicit solution of this differential equation.

Show that $3 x^{2} + 6 x y + 6 x - 2 y^{3} + 18 y - 4 = 0$ is an implicit solution of $\frac{d y}{d x} = \frac{x + y + 1}{y^{2} - x - 3}$ .

Solution

Use implicit differentiation to find $\frac{d y}{d x}$ . Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ :

\frac{d}{d x} [3 x^{2} + 6 x y + 6 x - 2 y^{3} + 18 y] - \frac{d}{d x} (4) = 0

The product rule handles the $6 x y$ term and the chain rule handles the $2 y^{3}$ term:

6 x + \underset{\frac{d}{d x} (6 x y)}{\underset{⏟}{6 y + 6 x \frac{d y}{d x}}} + 6 - \underset{\frac{d}{d x} (2 y^{3})}{\underset{⏟}{2 \cdot 3 y^{2} \frac{d y}{d x}}} + 18 \frac{d y}{d x} = 0

Collect the terms containing $\frac{d y}{d x}$ on one side and the rest on the other:

(6 x - 6 y^{2} + 18) \frac{d y}{d x} + 6 (x + y + 1) = 0

Solving for the derivative and simplifying the sign,

\frac{d y}{d x} = - \frac{x + y + 1}{x - y^{2} + 3} = \frac{x + y + 1}{y^{2} - x - 3}

which is the original differential equation. Notice that the last step multiplied numerator and denominator by $- 1$ : $- (x - y^{2} + 3) = y^{2} - x - 3$ .

General, Particular, and Singular Solutions

A solution of a differential equation may be one of three types: general, particular, or singular.

An ordinary differential equation of order $n$ has, in general, a solution containing $n$ arbitrary constants. We call such a solution the general solution.

For example,

y = c_{1} \sin x + c_{2} \cos x

where $c_{1}$ and $c_{2}$ are arbitrary constants is the general solution of y'' + y = 0. The count matches: the equation has order $2$ and the general solution carries $2$ constants.

A particular solution is a solution obtained from the general solution by giving the constants particular values.

Taking $c_{1} = 3$ and $c_{2} = - 1$ above gives the particular solution $y = 3 \sin x - \cos x$ .

A singular solution is a solution without arbitrary constants that cannot be derived from the general solution by assigning specific values to the constants. Such solutions arise only in nonlinear differential equations.

The equation \left(y'\right)^2 - 4y = 0 has general solution $y = (x + c)^{2}$ . Show that $y = 0$ is also a solution but is not a member of that family.

Solution

For $y = 0$ we have y' = 0, so \left(y'\right)^2 - 4y = 0 - 0 = 0. It is a solution.

Now look for a constant $c$ with $(x + c)^{2} = 0$ for every $x$ . No such $c$ exists, since $(x + c)^{2}$ vanishes at the single point $x = - c$ and nowhere else. Therefore $y = 0$ is a singular solution: it satisfies the equation but is not recoverable from the general solution.

Geometrically, $y = 0$ is the envelope of the family of parabolas $y = (x + c)^{2}$ , touching each one at its vertex. Envelopes are the usual source of singular solutions.

Exercises

Classify each equation by order, degree, and linearity.

\displaystyle{x^2 y'' + x y' + \left(x^2 - 4\right)y = 0}
\displaystyle{y' = y^2 + x}
\displaystyle{\left(y'''\right)^2 + y y' = 0}

Solution

Order $2$ , degree $1$ , linear and homogeneous. The coefficients $x^{2}$ , $x$ , and $x^{2} - 4$ depend on the independent variable only, which is allowed. This is Bessel's equation of order $2$ .
Order $1$ , degree $1$ , nonlinear because of the $y^{2}$ term. This is a Riccati equation.
Order $3$ , degree $2$ , nonlinear for two separate reasons: the third derivative is squared, and the product y y' appears.

Verify that $y = c_{1} e^{2 x} + c_{2} e^{- 2 x}$ is the general solution of y'' - 4y = 0, then find the particular solution satisfying $y (0) = 1$ and y'(0) = 6.

Solution

Differentiating twice gives y' = 2c_1 e^{2x} - 2c_2 e^{-2x} and y'' = 4c_1 e^{2x} + 4c_2 e^{-2x}, so

y'' - 4y = 4c_1 e^{2x} + 4c_2 e^{-2x} - 4c_1 e^{2x} - 4c_2 e^{-2x} = 0 .

The equation has order $2$ and the family carries $2$ constants, so this is the general solution.

Imposing the conditions at $x = 0$ :

\begin{aligned} y(0) &= c_1 + c_2 = 1 \\ y'(0) &= 2c_1 - 2c_2 = 6 \end{aligned}

The second equation gives $c_{1} - c_{2} = 3$ . Adding it to $c_{1} + c_{2} = 1$ gives $2 c_{1} = 4$ , so $c_{1} = 2$ and $c_{2} = - 1$ .

The particular solution is $y = 2 e^{2 x} - e^{- 2 x}$ .

Frequently Asked Questions

What is the difference between the order and the degree of a differential equation?

Order is the order of the highest derivative present. Degree is the exponent on that highest derivative, once the equation has been rationalized and cleared of fractions. In \left(y''\right)^3 + \left(y'\right)^8 = 0 the order is $2$ and the degree is $3$ . The exponent $8$ is irrelevant, because it sits on a lower-order derivative.

Can an equation be linear in

y

but contain complicated functions of

x

Yes, and this is the most common misunderstanding about linearity. The equation y'' + e^{x^2} y' + \ln(x)\, y = \tan x is perfectly linear. Linearity is a condition on the dependent variable $y$ and its derivatives, not on the independent variable. Coefficients and forcing terms may be arbitrarily messy functions of $x$ .

Why does linearity matter so much?

Linear equations obey the superposition principle: if $y_{1}$ and $y_{2}$ solve a homogeneous linear equation, then so does $c_{1} y_{1} + c_{2} y_{2}$ . That single fact makes the entire theory work, because it lets you build the general solution out of a finite set of independent building blocks. Nonlinear equations have no such structure, which is why they are handled case by case.

Is y'' + y = 0 homogeneous, and is y'' + y = 5 too?

The first is homogeneous, since the right-hand side is identically zero. The second is non-homogeneous: a nonzero constant counts as a nonzero forcing function. Only $f (x) \equiv 0$ qualifies as homogeneous.

When would I settle for an implicit solution?

Whenever solving for $y$ is impossible or would obscure the answer. The relation $3 x^{2} + 6 x y + 6 x - 2 y^{3} + 18 y - 4 = 0$ cannot be untangled into a clean formula $y = ϕ (x)$ , but it fully determines the solution curves. Separable and exact equations very often produce their answers in implicit form, and it is standard practice to leave them that way.

Does every differential equation have a singular solution?

No. Singular solutions arise only in nonlinear equations. A linear equation's general solution captures every solution there is, so nothing can be left over. In nonlinear equations the general solution can have an envelope, a curve tangent to every member of the family, and that envelope is itself a solution that no choice of constant reproduces.

How do I know how many arbitrary constants to expect?

An $n$ th order equation generally has a general solution with exactly $n$ arbitrary constants, one for each integration you would need to perform to undo the $n$ derivatives. A first-order equation has one, a second-order equation has two. If your answer has the wrong number of constants, you have almost certainly dropped a solution or introduced a redundant one.

Quick Reference

Order of a Differential Equation

The General $n$ th Order Form

Degree of a Differential Equation

Linear vs. Nonlinear, Homogeneous vs. Non-homogeneous

Homogeneous and Non-homogeneous

Linearity for Partial Differential Equations

Explicit and Implicit Solutions

Implicit Solutions

General, Particular, and Singular Solutions

Exercises

Frequently Asked Questions

Quick Links

Social Media

Classification of Differential Equations

Quick Reference

Order of a Differential Equation

The General n th Order Form

Degree of a Differential Equation

Linear vs. Nonlinear, Homogeneous vs. Non-homogeneous

Homogeneous and Non-homogeneous

Linearity for Partial Differential Equations

Explicit and Implicit Solutions

Implicit Solutions

General, Particular, and Singular Solutions

Exercises

Frequently Asked Questions

Quick Links

Social Media

The General $n$ th Order Form