
# 3.1: Basic Concepts

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## Basic Concepts

A $$\textcolor{blue}{\mbox{differential equation}}$$ is an equation that contains one or more derivatives of an unknown function. The $$\textcolor{blue}{\mbox{order}}$$ of a differential equation is the order of the highest derivative that it contains. A differential equation is an $$\textcolor{blue}{\mbox{ordinary differential equation}}$$ if it involves an unknown function of only one variable, or a $$\textcolor{blue}{\mbox{partial differential equation}}$$ if it involves partial derivatives of a function of more than one variable. For now we'll consider only ordinary differential equations, and we'll just call them $$\textcolor{blue}{\mbox{differential equations}}$$.

Throughout this chapter, all variables and constants are real unless it's stated otherwise. We'll usually use $$x$$ for the independent variable unless the independent variable is time; then we'll use $$t$$.

The simplest differential equations are first order equations of the form

${dy\over dx}=f(x) \nonumber$

or equivalently

$y'=f(x), \nonumber$

where $$f$$ is a known function of $$x$$. We already know from calculus how to find functions that satisfy this kind of equation. For example, if

$y'=x^3, \nonumber$

then

$y=\int x^3\, dx={x^4\over4}+c, \nonumber$

where $$c$$ is an arbitrary constant. If $$n>1$$ we can find functions $$y$$ that satisfy equations of the form

$\label{eq:3.2.1} y^{(n)}=f(x)$

by repeated integration. Again, this is a calculus problem.

Except for illustrative purposes in this section, there’s no need to consider differential equations like Equation \ref{eq:3.2.1}. We’ll usually consider differential equations that can be written as

$\label{eq:3.2.2} y^{(n)}=f(x,y,y', \dots,y^{(n-1)}),$

where at least one of the functions $$y$$, $$y'$$, …, $$y^{(n-1)}$$ actually appears on the right. Here are some examples:

\begin{array}{rcll} {dy\over dx}-x^2&=&0&\mbox{ (first order)}, \\ {dy\over dx}+2xy^2&=&-2&\mbox{ (first order)}, \\ {d^2y\over dx^2}+2 {dy\over dx}+y&=&2x&\mbox{ (second order)}, \\ xy'''+y^2&=&\sin x &\mbox{ (third order)}, \\ y^{(n)}+xy'+3y&=&x&\mbox{ (n-th order)}. \nonumber \end{array}

Although none of these equations is written as in Equation \ref{eq:3.2.2}, all of them can be written in this form:

$\begin{array}{rcl} y'&=&x^2, \\ y'&=&-2-2xy^2, \\ y''&=&2x-2y'-y, \\ y'''&=& \dfrac{\sin x-y^2}{x}, \\[4pt] y^{(n)}&=&x-xy'-3y. \end{array}\nonumber$

## Solutions of Differential Equations

A solution of a differential equation is a function that satisfies the differential equation on some open interval; thus, $$y$$ is a solution of Equation \ref{eq:3.2.2} if $$y$$ is $$n$$ times differentiable and

$y^{(n)}(x)=f(x,y(x),y'(x), \dots,y^{(n-1)}(x)) \nonumber$

for all $$x$$ in some open interval $$(a,b)$$. In this case, we also say that $$y$$ is a solution of Equation \ref{eq:3.2.2} on $$(a,b)$$. Functions that satisfy a differential equation at isolated points are not interesting. For example, $$y=x^2$$ satisfies

$xy'+x^2=3x \nonumber$

if and only if $$x=0$$ or $$x=1$$, but it is not a solution of this differential equation because it does not satisfy the equation on an open interval.

The graph of a solution of a differential equation is a solution curve. More generally, a curve $$C$$ is said to be an integral curve of a differential equation if every function $$y=y(x)$$ whose graph is a segment of $$C$$ is a solution of the differential equation. Thus, any solution curve of a differential equation is an integral curve, but an integral curve need not be a solution curve.

Example $$\PageIndex{1}$$

If $$a$$ is any positive constant, the circle

$\label{eq:3.2.3} x^2+y^2=a^2$

is an integral curve of

$\label{eq:3.2.4} y'=-{x\over y}.$

Solution

To see this, note that the only functions whose graphs are segments of Equation \ref{eq:3.2.3} are

$y_1=\sqrt{a^2-x^2} \quad \text{and} \quad y_2=-\sqrt{a^2-x^2}.\nonumber$

We leave it to you to verify that these functions both satisfy Equation \ref{eq:3.2.4} on the open interval $$(-a,a)$$. However, Equation \ref{eq:3.2.3} is not a solution curve of Equation \ref{eq:3.2.4}, since it is not the graph of a function.

Example $$\PageIndex{2}$$

Verify that

$\label{eq:3.2.5} y={x^2\over3}+{1\over x}$

is a solution of

$\label{eq:3.2.6} xy'+y=x^2$

on $$(0,\infty)$$ and on $$(-\infty,0)$$.

Solution

Substituting Equation \ref{eq:3.2.5} and

$y'={2x\over3} - {1\over x^2}\nonumber$

into Equation \ref{eq:3.2.6} yields

$xy'(x)+y(x)=x \left({2x\over3} - {1\over x^2}\right)+ \left({x^2\over3}+{1\over x}\right)=x^2\nonumber$

for all $$x\ne0$$. Therefore $$y$$ is a solution of Equation \ref{eq:3.2.6} on $$(-\infty,0)$$ and $$(0,\infty)$$. However, $$y$$ isn’t a solution of the differential equation on any open interval that contains $$x=0$$, since $$y$$ is not defined at $$x=0$$.

Figure $$\PageIndex{2}$$ shows the graph of Equation \ref{eq:3.2.5}. The part of the graph of Equation \ref{eq:3.2.5} on $$(0,\infty)$$ is a solution curve of Equation \ref{eq:3.2.6}, as is the part of the graph on $$(-\infty,0)$$.

Example $$\PageIndex{3}$$

Show that if $$c_1$$ and $$c_2$$ are constants then

$\label{eq:3.2.7} y=(c_1+c_2x)e^{-x}+2x-4$

is a solution of $\label{eq:3.2.8} y''+2y'+y=2x$ on $$(-\infty,\infty)$$.

Solution

Differentiating Equation \ref{eq:3.2.7} twice yields

$y'=-(c_1+c_2x)e^{-x}+c_2e^{-x}+2 \nonumber$

and

$y''=(c_1+c_2x)e^{-x}-2c_2e^{-x}, \nonumber$

so

\begin{align*} y''+2y'+y&=(c_1+c_2x)e^{-x}-2c_2e^{-x} + 2\left[-(c_1+c_2x)e^{-x}+c_2e^{-x}+2\right] +(c_1+c_2x)e^{-x}+2x-4 \\[4pt] &=(1-2+1)(c_1+c_2x)e^{-x}+(-2+2)c_2e^{-x} + 4+2x-4 \\[4pt] &=2x \end{align*}

for all values of $$x$$. Therefore $$y$$ is a solution of Equation \ref{eq:3.2.8} on $$(-\infty,\infty)$$.

Example $$\PageIndex{4}$$

Find all solutions of

$\label{eq:3.2.9} y^{(n)}=e^{2x}.$

Solution

Integrating Equation \ref{eq:3.2.9} yields

$y^{(n-1)}={e^{2x}\over2}+k_1, \nonumber$

where $$k_1$$ is a constant. If $$n\ge2$$, integrating again yields

$y^{(n-2)}={e^{2x}\over4}+k_1x+k_2. \nonumber$

If $$n\ge3$$, repeatedly integrating yields

$\label{eq:3.2.10} y={e^{2x}\over2^n}+k_1{x^{n-1}\over (n-1)!}+k_2{x^{n-2}\over (n-2)!}+\cdots+k_n,$

where $$k_1$$, $$k_2$$, …, $$k_n$$ are constants. This shows that every solution of Equation \ref{eq:3.2.9} has the form Equation \ref{eq:3.2.10} for some choice of the constants $$k_1$$, $$k_2$$, …, $$k_n$$. On the other hand, differentiating Equation \ref{eq:3.2.10} $$n$$ times shows that if $$k_1$$, $$k_2$$, …, $$k_n$$ are arbitrary constants, then the function $$y$$ in Equation \ref{eq:3.2.10} satisfies Equation \ref{eq:3.2.9}.

Since the constants $$k_1$$, $$k_2$$, …, $$k_n$$ in Equation \ref{eq:3.2.10} are arbitrary, so are the constants

${k_1\over (n-1)!},\, {k_2\over(n-2)!},\, \cdots, \, k_n. \nonumber$

Therefore Example $$\PageIndex{4}$$ actually shows that all solutions of Equation \ref{eq:3.2.9} can be written as

$y={e^{2x}\over2^n}+c_1+c_2x+\cdots+c_nx^{n-1}, \nonumber$

where we renamed the arbitrary constants in Equation \ref{eq:3.2.10} to obtain a simpler formula. As a general rule, arbitrary constants appearing in solutions of differential equations should be simplified if possible. You’ll see examples of this throughout the text.

## Initial Value Problems

In Example $$\PageIndex{4}$$ we saw that the differential equation $$y^{(n)}=e^{2x}$$ has an infinite family of solutions that depend upon the $$n$$ arbitrary constants $$c_1$$, $$c_2$$, …, $$c_n$$. In the absence of additional conditions, there’s no reason to prefer one solution of a differential equation over another. However, we’ll often be interested in finding a solution of a differential equation that satisfies one or more specific conditions. The next example illustrates this.

Example $$\PageIndex{5}$$

Find a solution of $y'=x^3 \nonumber$ such that $$y(1)=2$$.

Solution

At the beginning of this section we saw that the solutions of $$y'=x^3$$ are

$y={x^4\over4}+c. \nonumber$

To determine a value of $$c$$ such that $$y(1)=2$$, we set $$x=1$$ and $$y=2$$ here to obtain

$2=y(1)={1\over4}+c \nonumber$

so

$c={7\over4}. \nonumber$

Therefore the required solution is

$y={x^4+7\over4}. \nonumber$

Figure $$\PageIndex{2}$$ shows the graph of this solution. Note that imposing the condition $$y(1)=2$$ is equivalent to requiring the graph of $$y$$ to pass through the point $$(1,2)$$.

We can rewrite the problem considered in Example $$\PageIndex{5}$$ more briefly as

$y'=x^3,\quad y(1)=2.\nonumber$

We call this an initial value problem. The requirement $$y(1)=2$$ is an initial condition. Initial value problems can also be posed for higher order differential equations. For example,

$\label{eq:3.2.11} y'' - 2y'+3y=e^x, \quad y(0)=1, \quad y'(0)=2$

is an initial value problem for a second order differential equation where $$y$$ and $$y'$$ are required to have specified values at $$x=0$$. In general, an initial value problem for an $$n$$-th order differential equation requires $$y$$ and its first $$n-1$$ derivatives to have specified values at some point $$x_0$$. These requirements are the initial conditions.We’ll denote an initial value problem for a differential equation by writing the initial conditions after the equation, as in Equation \ref{eq:3.2.11}. For example, we would write an initial value problem for Equation \ref{eq:3.2.2} as

$\label{eq:3.2.12} y^{(n)}=f(x,y,y', \dots,y^{(n-1)}),\, y(x_0)=k_0,\, y'(x_0)=k_1,\, \dots,\, y^{(n-1)}=k_{n-1}.$

Consistent with our earlier definition of a solution of the differential equation in Equation \ref{eq:3.2.12}, we say that $$y$$ is a solution of the initial value problem Equation \ref{eq:3.2.12} if $$y$$ is $$n$$ times differentiable and

$y^{(n)}(x)=f(x,y(x),y'(x), \dots,y^{(n-1)}(x))\nonumber$

for all $$x$$ in some open interval $$(a,b)$$ that contains $$x_0$$, and $$y$$ satisfies the initial conditions in Equation \ref{eq:3.2.12}. The largest open interval that contains $$x_0$$ on which $$y$$ is defined and satisfies the differential equation is the interval of validity of $$y$$.

Example $$\PageIndex{6}$$

In Example $$\PageIndex{5}$$ we saw that

$\label{eq:3.2.13} y={x^4+7\over4}$

is a solution of the initial value problem

$y'=x^3,\quad y(1)=2.\nonumber$

Since the function in Equation \ref{eq:3.2.13} is defined for all $$x$$, the interval of validity of this solution is $$(-\infty,\infty)$$.

Example $$\PageIndex{7}$$

In Example $$\PageIndex{2}$$ we verified that

$\label{eq:3.2.14} y={x^2\over3}+{1\over x}$

is a solution of

$xy'+y=x^2 \nonumber$

on $$(0,\infty)$$ and on $$(-\infty,0)$$. By evaluating Equation \ref{eq:3.2.14} at $$x=\pm1$$, you can see that Equation \ref{eq:3.2.14} is a solution of the initial value problems

$\label{eq:3.2.15} xy'+y=x^2,\quad y(1)={4\over3}$

and

$\label{eq:3.2.16} xy'+y=x^2,\quad y(-1)=-{2\over3}.$

The interval of validity of Equation \ref{eq:3.2.14} as a solution of Equation \ref{eq:3.2.15} is $$(0,\infty)$$, since this is the largest interval that contains $$x_0=1$$ on which Equation \ref{eq:3.2.14} is defined. Similarly, the interval of validity of Equation \ref{eq:3.2.14} as a solution of Equation \ref{eq:3.2.16} is $$(-\infty,0)$$, since this is the largest interval that contains $$x_0=-1$$ on which Equation \ref{eq:3.2.14} is defined.

## Free Fall Under Constant Gravity

The term initial value problem originated in problems of motion where the independent variable is $$t$$ (representing elapsed time), and the initial conditions are the position and velocity of an object at the initial (starting) time of an experiment.

Example $$\PageIndex{8}$$

An object falls under the influence of gravity near Earth’s surface, where it can be assumed that the magnitude of the acceleration due to gravity is a constant $$g$$.

1. Construct a mathematical model for the motion of the object in the form of an initial value problem for a second-order differential equation, assuming that the altitude and velocity of the object at time $$t=0$$ are known. Assume that gravity is the only force acting on the object.
2. Solve the initial value problem derived above to obtain the altitude as a function of time.

Solution a

Let $$y(t)$$ be the altitude of the object at time $$t$$. Since the acceleration of the object has constant magnitude $$g$$ and is in the downward (negative) direction, $$y$$ satisfies the second order equation

$y''=-g, \nonumber$

where the prime now indicates differentiation with respect to $$t$$. If $$y_0$$ and $$v_0$$ denote the altitude and velocity when $$t=0$$, then $$y$$ is a solution of the initial value problem

$\label{eq:3.2.17} y''=-g,\quad y(0)=y_0,\quad y'(0)=v_0.$

Solution b

Integrating Equation \ref{eq:3.2.17} twice yields

\begin{aligned} y'&=-gt+c_1, \\ y&=-{gt^2\over2}+c_1t+c_2.\end{aligned}

Imposing the initial conditions $$y(0)=y_0$$ and $$y'(0)=v_0$$ in these two equations shows that $$c_1=v_0$$ and $$c_2=y_0$$. Therefore the solution of the initial value problem Equation \ref{eq:3.2.17} is

$y=- {gt^2\over2}+v_0t+y_0. \nonumber$

Trench, William F., "Elementary Differential Equations" (2013). Faculty Authored and Edited Books & CDs. 8.
https://digitalcommons.trinity.edu/mono/8