1.2 Basic Concepts

A differential equation is an equation that contains one or more derivatives of an unknown function. The order of a differential equation is the order of the highest derivative that it contains. A differential equation is an ordinary differential equation if it involves an unknown function of only one variable, or a 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 differential equations.

Throughout this text, 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

\[ {dy\over dx}=f(x) \mbox{\quad or, equivalently, \quad}y'=f(x), \]

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

\[ y'=x^3, \]

then

\[ y=\int x^3\, dx={x^4\over4}+c, \]

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

\begin{equation} y^{(n)}=f(x) \tag{1.2.1}\end{equation}

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 (1.2.1).We’ll usually consider differential equations that can be written as

\begin{equation} y^{(n)}=f(x,y,y', \dots,y^{(n-1)}), \tag{1.2.2}\end{equation}

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

\[ \begin{array}{rcll} \dst{dy\over dx}-x^2&=&0&\mbox{ (first order)}, \\[6pt] \dst{dy\over dx}+2xy^2&=&-2&\mbox{ (first order)}, \\[6pt] \dst{d^2y\over dx^2}+2\dst{dy\over dx}+y&=&2x&\mbox{ (second order)}, \\[6pt] xy'''+y^2&=&\sin x &\mbox{ (third order)}, \\[6pt] y^{(n)}+xy'+3y&=&x&\mbox{ ($n$-th order)}. \end{array} \]

Although none of these equations is written as in (1.2.2), all of them can be written in this form:

\[ \begin{array}{rcl} y'&=&x^2, \\ y'&=&-2-2xy^2, \\ y''&=&2x-2y'-y, \\ y'''&=&\dst{\sin x-y^2\over x}, \\[6pt] y^{(n)}&=&x-xy'-3y. \end{array} \]

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 (1.2.2) if \(y\) is \(n\) times differentiable and

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

for all \(x\) in some open interval \((a,b)\). In this case, we also say that \(y\) is a solution of \(\eqref{eq:1.2.2}\) on \((a,b)\). Functions that satisfy a differential equation at isolated points are not interesting. For example, \(y=x^2\) satisfies

\[ xy'+x^2=3x \]

if and only if \(x=0\) or \(x=1\), but it’s 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 1.2.1

If \(a\) is any positive constant, the circle

\begin{equation} x^2+y^2=a^2 \tag{1.2.3}\end{equation}

is an integral curve of

\begin{equation} y'=-{x\over y}. \tag{1.2.4}\end{equation}

To see this, note that the only functions whose graphs are segments of (1.2.3) are

\[ y_1=\sqrt{a^2-x^2}\mbox{\quad and \quad} y_2=-\sqrt{a^2-x^2}. \]

We leave it to you to verify that these functions both satisfy (1.2.4) on the open interval \((-a,a)\). However, (1.2.3) is not a solution curve of (1.2.4), since it’s not the graph of a function.

Example 1.2.2

Verify that

\begin{equation} y={x^2\over3}+{1\over x} \tag{1.2.5}\end{equation}

is a solution of

\begin{equation} xy'+y=x^2 \tag{1.2.6}\end{equation}

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

Solution Substituting (1.2.5) and

\[ y'={2x\over3} - {1\over x^2} \]

into (1.2.6) yields

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

for all \(x\ne0\). Therefore \(y\) is a solution of (1.2.6) 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 1.2.1 shows the graph of (1.2.5). The part of the graph of (1.2.5) on \((0,\infty)\) is a solution curve of (1.2.6), as is the part of the graph on \((-\infty,0)\).

Example 1.2.3

Show that if \(c_1\) and \(c_2\) are constants then

\begin{equation} y=(c_1+c_2x)e^{-x}+2x-4 \tag{1.2.7}\end{equation}

is a solution of

\begin{equation} y''+2y'+y=2x \tag{1.2.8}\end{equation}

on \((-\infty,\infty)\).

y=x^23+1x
Figure 1.2.1. \(y=\dst{\frac{x^{2}}{3}+\frac{1}{x}}\)

Solution Differentiating (1.2.7) twice yields

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

and

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

so

\begin{eqnarray*} 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 \\ &=&(1-2+1)(c_1+c_2x)e^{-x}+(-2+2)c_2e^{-x} \\ &&+4+2x-4=2x \end{eqnarray*}

for all values of \(x\). Therefore \(y\) is a solution of (1.2.8) on \((-\infty,\infty)\).

Example 1.2.4

Find all solutions of

\begin{equation} y^{(n)}=e^{2x}. \tag{1.2.9}\end{equation}

Solution Integrating (1.2.9) yields

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

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

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

If \(n\ge3\), repeatedly integrating yields

\begin{equation} y={e^{2x}\over2^n}+k_1{x^{n-1}\over (n-1)!}+k_2{x^{n-2}\over (n-2)!}+\cdots+k_n, \tag{1.2.10}\end{equation}

where \(k_1\), \(k_2\), …, \(k_n\) are constants. This shows that every solution of (1.2.9) has the form (1.2.10) for some choice of the constants \(k_1\), \(k_2\), …, \(k_n\). On the other hand, differentiating (1.2.10) \(n\) times shows that if \(k_1\), \(k_2\), …, \(k_n\) are arbitrary constants, then the function \(y\) in (1.2.10) satisfies (1.2.9).

Since the constants \(k_1\), \(k_2\), …, \(k_n\) in (1.2.10) are arbitrary, so are the constants

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

Therefore Example 1.2.4 actually shows that all solutions of (1.2.9) can be written as

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

where we renamed the arbitrary constants in (1.2.10) 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 1.2.4 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 1.2.5

Find a solution of

\[ y'=x^3 \]

such that \(y(1)=2\).

Solution At the beginning of this section we saw that the solutions of \(y'=x^3\) are

\[ y={x^4\over4}+c. \]

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

\[ 2=y(1)={1\over4}+c,\mbox{\quad so \quad}c={7\over4}. \]

Therefore the required solution is

\[ y={x^4+7\over4}. \]

Figure 1.2.2 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 1.2.5 more briefly as

\[ y'=x^3,\quad y(1)=2. \]

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,

\begin{equation} y'' - 2y'+3y=e^x, \quad y(0)=1, \quad y'(0)=2 \tag{1.2.11}\end{equation}

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.

y=x^2+74
Figure 1.2.2. \(y=\dst{\frac{x^{2}+7}{4}}\)

We’ll denote an initial value problem for a differential equation by writing the initial conditions after the equation, as in (1.2.11). For example, we would write an initial value problem for (1.2.2) as

\begin{equation} 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}. \tag{1.2.12}\end{equation}

Consistent with our earlier definition of a solution of the differential equation in (1.2.12), we say that \(y\) is a solution of the initial value problem (1.2.12) if \(y\) is \(n\) times differentiable and

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

for all \(x\) in some open interval \((a,b)\) that contains \(x_0\), and \(y\) satisfies the initial conditions in (1.2.12). 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 1.2.6

In Example 1.2.5 we saw that

\begin{equation} y={x^4+7\over4} \tag{1.2.13}\end{equation}

is a solution of the initial value problem

\[ y'=x^3,\quad y(1)=2. \]

Since the function in (1.2.13) is defined for all \(x\), the interval of validity of this solution is \((-\infty,\infty)\).

Example 1.2.7

In Example 1.2.2 we verified that

\begin{equation} y={x^2\over3}+{1\over x} \tag{1.2.14}\end{equation}

is a solution of

\[ xy'+y=x^2 \]

on \((0,\infty)\) and on \((-\infty,0)\). By evaluating (1.2.14) at \(x=\pm1\), you can see that (1.2.14) is a solution of the initial value problems

\begin{equation} xy'+y=x^2,\quad y(1)={4\over3} \tag{1.2.15}\end{equation}

and

\begin{equation} xy'+y=x^2,\quad y(-1)=-{2\over3}. \tag{1.2.16}\end{equation}

The interval of validity of (1.2.14) as a solution of (1.2.15) is \((0,\infty)\), since this is the largest interval that contains \(x_0=1\) on which (1.2.14) is defined. Similarly, the interval of validity of (1.2.14) as a solution of (1.2.16) is \((-\infty,0)\), since this is the largest interval that contains \(x_0=-1\) on which (1.2.14) 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 1.2.8

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 in (a) 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

\[ y''=-g, \]

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

\begin{equation} y''=-g,\quad y(0)=y_0,\quad y'(0)=v_0. \tag{1.2.17}\end{equation}

Solution (b) Integrating (1.2.17) twice yields

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

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 (1.2.17) is

\[ y=- {gt^2\over2}+v_0t+y_0. \]

1.2 Exercises

  1. Find the order of the equation.

    (a) \(\dst{d^2y\over dx^2}+2 {dy\over dx}\ {d^3y\over dx^3}+x=0\)(b) \(y''-3y'+2y=x^7\)
    (c) \(y'-y^7=0\)(d) \(y''y-(y')^2=2\)
    Show answer

    (a)  \(3\)  (b)  \(2\)   (c)  \(1\)  (d)  \(2\)

  2. Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function.

    1. \(y=ce^{2x}; \quad y'=2y\)

    2. \(y=\dst{x^2\over3} +{c\over x}; \quad xy'+y=x^2\)

    3. \(y=\dst{1\over2}+ce^{-x^2}; \quad y'+2xy=x\)

    4. \(y=(1+ce^{-x^2/2}); (1-ce^{-x^2/2})^{-1} \quad 2y'+x(y^2-1)=0\)

    5. \(y=\dst{\tan\left( {x^3\over3}+c\right)}; \quad y'=x^2(1+y^2)\)

    6. \(y=(c_1+c_2x)e^x+\sin x+x^2; \quad y''-2y'+y=-2 \cos x+x^2-4x+2\)

    7. \(y=c_1e^x+c_2x+\dst{2\over x}; \quad (1-x)y''+xy'- y=4(1-x-x^2)x^{-3}\)

    8. \(y=x^{-1/2}(c_1\sin x+c_2 \cos x)+4x+8\); \(x^2y''+xy'+\dst{\left(x^2-{1\over4}\right)}y=4x^3+8x^2+3x-2\)

  3. Find all solutions of the equation.

    (a)  \(y'=-x\)(b)  \(y'=-x \sin x\)
    (c)  \(y'=x \ln x\)(d)  \(y''=x \cos x\)
    (e)  \(y''=2xe^x\)(f)  \(y''=2x+\sin x+e^x\)
    (g)  \(y'''=-\cos x\)(h)  \(y'''=-x^2+e^x\)
    (i)  \(y'''=7e^{4x}\)
    Show answer

    (a) \(y=-\dst{\frac{x^{2}}{2}}+c\)   (b) \(y=x\cos x-\sin x+c\)

       (c) \(y=\dst{\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+c}\) (d) \(y=-x\cos x+2\sin x+c_{1}+c_{2}x\)

         (e) \(y=(2x-4)e^{x}+c_{1}+c_{2}x\)   (f) \(y=\dst{\frac{x^{3}}{3}}-\sin x+e^{x}+c_{1}+c_{2}x\)

         (g) \(y=\sin x+c_{1}+c_{2}x+c_{3}x^{2}\) (h) \(y=-\dst{\frac{x^{5}}{60}}+e^{x}+c_{1}+c_{2}x+c_{3}x^{2}\)

         (i) \(y=\dst{\frac{7}{64}}e^{4x}+c_{1}+c_{2}x+c_{3}x^{2}\)

  4. Solve the initial value problem.

    1. \(y'=-xe^x, \quad y(0)=1\)

    2. \(\dst{y'=x \sin x^2, \quad y\left({\sqrt{\pi\over2}}\right)=1}\)

    3. \(y'=\tan x, \quad y(\pi/4)=3\)

    4. \(y''=x^4, \quad y(2)=-1, \quad y'(2)=-1\)

    5. \(y''=xe^{2x}, \quad y(0)=7, \quad y'(0)=1\)

    6. \(y''=- x \sin x, \quad y(0)=1, \quad y'(0)=-3\)

    7. \(y'''=x^2e^x, \quad y(0)=1, \quad y'(0)=-2, \quad y''(0)=3\)

    8. \(y'''=2+\sin 2x, \quad y(0)=1, \quad y'(0)=-6, \quad y''(0)=3\)

    9. \(y'''=2x+1, \quad y(2)=1, \quad y'(2)=-4, \quad y''(2)=7\)

    Show answer

    (a) \(y=-(x-1)e^{x}\) (b) \(y=1-\dst{\frac{1}{2}}\cos x^{2}\) (c) \(y=3-\ln(\sqrt{2}\cos x)\)

       (d) \(y=\dst{-\frac{47}{15}-\frac{37}{5}(x-2)+\frac{x^{5}}{30}}\) (e) \(y=\dst{\frac{1}{4}xe^{2x}-\frac{1}{4}e^{2x}+\frac{29}{4}}\)

         (f) \(y=x\sin x+2\cos x-3x-1\) (g) \(y=(x^{2}-6x+12)e^{x}+\dst{\frac{x^{2}}{2}}-8x-11\)

         (h) \(y=\dst{\frac{x^{3}}{3}+\frac{\cos 2x}{6}+\frac{7}{4}x^{2}-6x+\frac{7}{8}}\) (i) \(y=\dst{\frac{x^{4}}{12}+\frac{x^{3}}{6}+\frac{1}{2}(x-2)^{2} -\frac{26}{3}(x-2)-\frac{5}{3}}\)

  5. Verify that the function is a solution of the initial value problem.

    1. \(y=x\cos x; \quad y'=\cos x-y\tan x, \quad y(\pi/4)=\dst{\pi\over4\sqrt{2}}\)

    2. \(\dst{y={1+2\ln x\over x^2}+{1\over2}; \quad y'={x^2-2x^2y+2\over x^3}, \quad y(1)={3\over2}}\)

    3. \(y=\dst{\tan\left({x^2\over2}\right)}; \quad y'=x(1+y^2), \quad y(0)=0\)

    4. \(\dst{y={2\over x-2}; \quad y'={-y(y+1)\over x}}, \quad y(1)=-2\)

  6. Verify that the function is a solution of the initial value problem.

    1. \(y=x^2(1+\ln x); \quad y''=\dst{3xy'-4y\over x^2}, \quad y(e)=2e^2, \quad y'(e)=5e\)

    2. \(y=\dst{x^2\over3}+x-1; \quad y''=\dst{x^2-xy'+y+1\over x^2}, \quad y(1)=\dst{1\over3}, \quad y'(1)=\dst{5\over3}\)

    3. \(y=(1+x^2)^{-1/2}; \quad y''=\dst{(x^2-1)y-x(x^2+1)y'\over (x^2+1)^2}, \quad y(0)=1,\)

      \(y'(0)=0\)

    4. \(y=\dst{x^2\over 1-x}; \quad y''=\dst{2(x+y)(xy'-y)\over x^3}, \quad y(1/2)=1/2, \quad y'(1/2)=3\)

  7. Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take \(g=32\) ft/sec\(^2\).

    1. Find the highest altitude attained by the object.

    2. Determine how long it takes for the object to fall to the ground.

    Show answer

    (a)  \(576\) ft  (b)  \(10\) s

  8. Let \(a\) be a nonzero real number.

    1. Verify that if \(c\) is an arbitrary constant then

      \[ y=(x-c)^a \tag*{\rm (A)} \]

      is a solution of

      \[ y'=ay^{(a-1)/a} \tag*{\rm (B)} \]

      on \((c,\infty)\).

    2. Suppose \(a<0\) or \(a>1\). Can you think of a solution of (B) that isn’t of the form (A)?

    Show answer

    (b)  \(y=0\)

  9. Verify that

    \[ y= \left\{ \begin{array}{cl} e^x-1,& x \ge 0, \\[6pt] 1-e^{-x},& x < 0, \end{array}\right. \]

    is a solution of

    \[ y'=|y|+1 \]

    on \((-\infty,\infty)\). Hint: Use the definition of derivative at \(x=0\).

    1. Verify that if \(c\) is any real number then

      \[ y=c^2+cx+2c+1 \tag*{\rm (A)} \]

      satisfies

      \[ y'={-(x+2)+\sqrt{x^2+4x+4y}\over2} \tag*{\rm (B)} \]

      on some open interval. Identify the open interval.

    2. Verify that

      \[ y_1={-x(x+4)\over4} \]

      also satisfies (B) on some open interval, and identify the open interval. (Note that \(y_1\) can’t be obtained by selecting a value of \(c\) in (A).)

    Show answer

    (a)  \((-2c-2,\infty)\)  \((-\infty,\infty)\)