
# 2.2: Classification of Differential Equations

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Recall that a differential equation is an equation (has an equal sign) that involves derivatives. Just as biologists have a classification system for life, mathematicians have a classification system for differential equations. We can place all differential equation into two types: ordinary differential equation and partial differential equations.

Examples $$\PageIndex{1}$$

$\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = 3x\; \sin \; y$

is an ordinary differential equation since it does not contain partial derivatives. While

$\dfrac{\partial y}{\partial t} + x \dfrac{\partial y}{\partial x} = \dfrac{x+t}{x-t}$

is a partial differential equation, since $$y$$ is a function of the two variables $$x$$ and $$t$$ and partial derivatives are present.

In this course we will focus on only ordinary differential equations.

### Order

Another way of classifying differential equations is by order. Any ordinary differential equation can be written in the form

$F(x,y,y',y'',...,y^{(0)})=0$

by setting everything equal to zero. The order of a differential equation is the highest derivative that appears in the above equation.

Examples $$\PageIndex{2}$$

$\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} = 3x\, \sin \; y$

is a second order differential equation, since a second derivative appears in the equation.

$3y^4y''' - x^3y' + e^{xy}y = 0$

is a third order differential equation.

Once we have written a differential equation in the form

$F(x,y,y',y'',...,y^{(n)}) = 0$

we can talk about whether a differential equation is linear or not. We say that the differential equation above is a linear differential equation if

$\dfrac{\partial F}{\partial y^{(i)} \partial y^{(j)} } = 0$

for all $$i$$ and $$j$$.  Any linear ordinary differential equation of degree $$n$$ can be written as

$a_0(x)y^{(n)} + a_1(x)y^{(n-1)} +\, ... + a_{n-1}(x)y' + a_n(x)y = g(x) .$

Examples $$\PageIndex{3}$$

$3x^2y'' + 2\ln \, (x)y' + e^x \, y = 3x\, \text{cos} \, x$

is a second order linear ordinary differential equation.

$4yy''' - x^3y' + \text{cos}\, y = e^{2x}$

is not a linear differential equation because of the $$4yy'''$$ and the $$\cos y$$ terms.

Nonlinear differential equations are often very difficult or impossible to solve. One approach getting around this difficulty is to linearize the differential equation.

Example $$\PageIndex{4}$$: Linearization

$y'' + 2y' + e^y = x$

is nonlinear because of the $$e^y$$ term. However, the Taylor expansion of the exponetial function

$e^y = 1 + y + \frac{y^2}{2} + \frac{y^3}{6} + \, ...$

can be approximated by the first two terms

$e^y \approx 1 + y.$

We instead solve the much easier linear differential equation

$y'' + 2y' + 1 + y = x.$

We say that a function $$f(x)$$ is a solution to a differential equation if plugging in $$f(x)$$ into the equation makes the equation equal.

Example $$\PageIndex{5}$$

Show that

$f(x) = x + e^{2x}$

is a solution to

$y'' - 2y' = -2.$

Solution

Taking derivatives:

$f'(x) = 1 + 2e^{2x} , f''(x) = 4e^{2x}.$

Now plug in to get

$4e^{2x} - 2(1 + 2e^{2x}) = 4e^{2x} - 2 - 4e^{2x} = -2 .$

Hence it is a solution.

Two questions that will be asking repeatedly of a differential equation course are

1. Does there exist a solution to the differential equation?
2. Is the solution given unique?

In the example above, the answer to the first question is yes since we verified that

$f(x)=x+e^{2x}$

is a solution.  However, the answer to the second question is no. It can be verified that

$s(x) = 4 + x$

is also a solution.

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.