# 2.2: Classification of Differential Equations

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.

- A
**partial**differential equation is a differential equation that**involves**partial derivatives. - An
**ordinary**differential equation is a differential equation that**does not**involve partial derivatives.

Examples

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

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

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

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

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

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: 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

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

- Does there exist a solution to the differential equation?
- 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.

### Contributors

- Larry Green (Lake Tahoe Community College)