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1 First-Order Differential Equations

1.1 Differential Equations and Mathematical Models

The laws of the universe are written in the language of mathematics. Algebra is sufficient to solve many static problems, but the most interesting natural phenomena involve change and are described by equations that relate changing quantities.

Because the derivative dx/dt=f'(t) $d x / d t = f ′ (t)$ of the function f is the rate at which the quantity x=f(t) $x = f (t)$ is changing with respect to the independent variable t, it is natural that equations involving derivatives are frequently used to describe the changing universe. An equation relating an unknown function and one or more of its derivatives is called a differential equation.

Example 1

The differential equation

d x d t = x 2 + t 2

$\frac{d x}{d t} = x^{2} + t^{2}$

involves both the unknown function x(t) and its first derivative x'(t)=dx/dt $x ′ (t) = d x / d t$ . The differential equation

d 2 y d x 2 + 3 d y d x + 7 y = 0

$\frac{d^{2} y}{d x^{2}} + 3 \frac{d y}{d x} + 7 y = 0$

involves the unknown function y of the independent variable x and the first two derivatives y' $y ′$ and y'' $y ″$ of y.

The study of differential equations has three principal goals:

To discover the differential equation that describes a specified physical situation.
To find—either exactly or approximately—the appropriate solution of that equation.
To interpret the solution that is found.

In algebra, we typically seek the unknown numbers that satisfy an equation such as x3+7x2−11x+41=0 $x^{3} + 7 x^{2} - 11 x + 41 = 0$ . By contrast, in solving a differential equation, we are challenged to find the unknown functions y=y(x) $y = y (x)$ for which an identity such as y'(x)=2xy(x) $y ′ (x) = 2 x y (x)$ —that is, the differential equation

d y d x = 2 x y

$\frac{d y}{d x} = 2 x y$

—holds on some interval of real numbers. Ordinarily, we will want to find all solutions of the differential equation, if possible.

Example 2

If C is a constant and

y (x) = C e x 2,

$y (x) = C e^{x^{2}},$ (1)

then

d y d x = C (2 x e x 2) = (2 x) (C e x 2) = 2 x y .

$\frac{d y}{d x} = C (2 x e^{x^{2}}) = (2 x) (C e^{x^{2}}) = 2 x y .$

Thus every function y(x) of the form in Eq. (1) satisfies—and thus is a solution of—the differential equation

d y d x = 2 x y

$\frac{d y}{d x} = 2 x y$ (2)

for all x. In particular, Eq. (1) defines an infinite family of different solutions of this differential equation, one for each choice of the arbitrary constant C. By the method of separation of variables (Section 1.4) it can be shown that every solution of the differential equation in (2) is of the form in Eq. (1).

Differential Equations and Mathematical Models

The following three examples illustrate the process of translating scientific laws and principles into differential equations. In each of these examples the independent variable is time t, but we will see numerous examples in which some quantity other than time is the independent variable.

Example 3

Rate of cooling Newton’s law of cooling may be stated in this way: The time rate of change (the rate of change with respect to time t) of the temperature T(t) of a body is proportional to the difference between T and the temperature A of the surrounding medium (Fig. 1.1.1). That is,

d T d t = - k (T - A),

$\frac{d T}{d t} = - k (T - A),$ (3)

where k is a positive constant. Observe that if T>A $T > A$ , then dT/dt<0 $d T / d t < 0$ , so the temperature is a decreasing function of t and the body is cooling. But if T<A $T < A$ , then dT/dt>0 $d T / d t > 0$ , so that T is increasing.

Thus the physical law is translated into a differential equation. If we are given the values of k and A, we should be able to find an explicit formula for T(t), and then—with the aid of this formula—we can predict the future temperature of the body.

FIGURE 1.1.1.

Newton’s law of cooling, Eq. (3), describes the cooling of a hot rock in water.

Example 4

Draining tank Torricelli’s law implies that the time rate of change of the volume V of water in a draining tank (Fig. 1.1.2) is proportional to the square root of the depth y of water in the tank:

d V d t = - k y \sqrt,

$\frac{d V}{d t} = - k \sqrt{y},$ (4)

where k is a constant. If the tank is a cylinder with vertical sides and cross-sectional area A, then V=Ay $V = A y$ , so dV/dt=A⋅(dy/dt) $d V / d t = A \cdot (d y / d t)$ . In this case Eq. (4) takes the form

d y d t = - h y \sqrt,

$\frac{d y}{d t} = - h \sqrt{y},$ (5)

where h=k/A $h = k / A$ is a constant.

FIGURE 1.1.2.

Torricelli’s law of draining, Eq. (4), describes the draining of a water tank.

Example 5

Population growth The time rate of change of a population P(t) with constant birth and death rates is, in many simple cases, proportional to the size of the population. That is,

d P d t = k P,

$\frac{d P}{d t} = k P,$ (6)

where k is the constant of proportionality.

Let us discuss Example 5 further. Note first that each function of the form

P (t) = C e k t

$P (t) = C e^{k t}$ (7)

is a solution of the differential equation

d P d t = k P

$\frac{d P}{d t} = k P$

in (6). We verify this assertion as follows:

P' (t) = C k e k t = k (C e k t) = k P (t)

$P^{′} (t) = C k e^{k t} = k (C e^{k t}) = k P (t)$

for all real numbers t. Because substitution of each function of the form given in (7) into Eq. (6) produces an identity, all such functions are solutions of Eq. (6).

Thus, even if the value of the constant k is known, the differential equation dP/dt=kP $d P / d t = k P$ has infinitely many different solutions of the form P(t)=Cekt $P (t) = C e^{k t}$ , one for each choice of the “arbitrary” constant C. This is typical of differential equations. It is also fortunate, because it may allow us to use additional information to select from among all these solutions a particular one that fits the situation under study.

Example 6

Population growth Suppose that P(t)=Cekt $P (t) = C e^{k t}$ is the population of a colony of bacteria at time t, that the population at time t=0 $t = 0$ (hours, h) was 1000, and that the population doubled after 1 h. This additional information about P(t) yields the following equations:

1000 = P (0) = C e 0 = C, 2000 = P (1) = C e k .

$\begin{array}{l} 1000 = P (0) = C e^{0} = C, \\ 2000 = P (1) = C e^{k} . \end{array}$

It follows that C=1000 $C = 1000$ and that ek=2 $e^{k} = 2$ , so k=ln 2≈0.693147 $k = \ln 2 \approx 0.693147$ . With this value of k the differential equation in (6) is

d P d t = (ln 2) P \approx (0.693147) P .

$\frac{d P}{d t} = (\ln 2) P \approx (0.693147) P .$

Substitution of k=ln 2 $k = \ln 2$ and C=1000 $C = 1000$ in Eq. (7) yields the particular solution

P (t) = 1000 e (ln 2) t = 1000 (e ln 2) t = 1000 \cdot 2 t \cdot \cdot (because e ln 2 = 2)

$\begin{array}{l} P (t) = 1000 e^{(\ln 2) t} = 1000 {(e^{\ln 2})}^{t} = 1000 \cdot 2^{t} & \cdot \cdot (because e^{\ln 2} = 2) \end{array}$

that satisfies the given conditions. We can use this particular solution to predict future populations of the bacteria colony. For instance, the predicted number of bacteria in the population after one and a half hours (when t=1.5 $t = 1.5$ ) is

P (1.5) = 1000 \cdot 2 3/2 \approx 2828 .

$P (1.5) = 1000 \cdot 2^{3/2} \approx 2828 .$

The condition P(0)=1000 $P (0) = 1000$ in Example 6 is called an initial condition because we frequently write differential equations for which t=0 $t = 0$ is the “starting time.” Figure 1.1.3 shows several different graphs of the form P(t)=Cekt $P (t) = C e^{k t}$ with k=ln 2 $k = \ln 2$ . The graphs of all the infinitely many solutions of dP/dt=kP $d P / d t = k P$ in fact fill the entire two-dimensional plane, and no two intersect. Moreover, the selection of any one point P0 $P_{0}$ on the P-axis amounts to a determination of P(0) $P (0)$ . Because exactly one solution passes through each such point, we see in this case that an initial condition P(0)=P0 $P (0) = P_{0}$ determines a unique solution agreeing with the given data.

FIGURE 1.1.3.

Graphs of P(t)=Cekt $P (t) = {C e}^{k t}$ with k=ln 2 $k = \ln 2$ .

Mathematical Models

Our brief discussion of population growth in Examples 5 and 6 illustrates the crucial process of mathematical modeling (Fig. 1.1.4), which involves the following:

The formulation of a real-world problem in mathematical terms; that is, the construction of a mathematical model.
The analysis or solution of the resulting mathematical problem.
The interpretation of the mathematical results in the context of the original real-world situation—for example, answering the question originally posed.

FIGURE 1.1.4.

The process of mathematical modeling.

In the population example, the real-world problem is that of determining the population at some future time. A mathematical model consists of a list of variables (P and t) that describe the given situation, together with one or more equations relating these variables (dP/dt=kP, P(0)=P0 $d P / d t = k P, P (0) = P_{0}$ ) that are known or are assumed to hold. The mathematical analysis consists of solving these equations (here, for P as a function of t). Finally, we apply these mathematical results to attempt to answer the original real-world question.

As an example of this process, think of first formulating the mathematical model consisting of the equations dP/dt=kP, P(0)=1000 $d P / d t = k P, P (0) = 1000$ , describing the bacteria population of Example 6. Then our mathematical analysis there consisted of solving for the solution function P(t)=1000e(ln 2)t=1000⋅2t $P (t) = 1000 e^{(\ln 2) t} = 1000 \cdot 2^{t}$ as our mathematical result. For an interpretation in terms of our real-world situation—the actual bacteria population—we substituted t=1.5 $t = 1.5$ to obtain the predicted population of P(1.5)≈2828 $P (1.5) \approx 2828$ bacteria after 1.5 hours. If, for instance, the bacteria population is growing under ideal conditions of unlimited space and food supply, our prediction may be quite accurate, in which case we conclude that the mathematical model is adequate for studying this particular population.

On the other hand, it may turn out that no solution of the selected differential equation accurately fits the actual population we’re studying. For instance, for no choice of the constants C and k does the solution P(t)=Cekt $P (t) = {C e}^{k t}$ in Eq. (7) accurately describe the actual growth of the human population of the world over the past few centuries. We must conclude that the differential equation dP/dt=kP $d P / d t = k P$ is inadequate for modeling the world population—which in recent decades has “leveled off” as compared with the steeply climbing graphs in the upper half (P>0 $P > 0$ ) of Fig. 1.1.3. With sufficient insight, we might formulate a new mathematical model including a perhaps more complicated differential equation, one that takes into account such factors as a limited food supply and the effect of increased population on birth and death rates. With the formulation of this new mathematical model, we may attempt to traverse once again the diagram of Fig. 1.1.4 in a counterclockwise manner. If we can solve the new differential equation, we get new solution functions to compare with the real-world population. Indeed, a successful population analysis may require refining the mathematical model still further as it is repeatedly measured against real-world experience.

But in Example 6 we simply ignored any complicating factors that might affect our bacteria population. This made the mathematical analysis quite simple, perhaps unrealistically so. A satisfactory mathematical model is subject to two contradictory requirements: It must be sufficiently detailed to represent the real-world situation with relative accuracy, yet it must be sufficiently simple to make the mathematical analysis practical. If the model is so detailed that it fully represents the physical situation, then the mathematical analysis may be too difficult to carry out. If the model is too simple, the results may be so inaccurate as to be useless. Thus there is an inevitable tradeoff between what is physically realistic and what is mathematically possible. The construction of a model that adequately bridges this gap between realism and feasibility is therefore the most crucial and delicate step in the process. Ways must be found to simplify the model mathematically without sacrificing essential features of the real-world situation.

Mathematical models are discussed throughout this book. The remainder of this introductory section is devoted to simple examples and to standard terminology used in discussing differential equations and their solutions.

Examples and Terminology

Example 7

If C is a constant and y(x)=1/(C−x) $y (x) = 1 / (C - x)$ , then

d y d x = 1 ( C - x ) 2 = y 2

$\frac{d y}{d x} = \frac{1}{{(C - x)}^{2}} = y^{2}$

if x≠C $x \neq C$ . Thus

y (x) = 1 C - x

$y (x) = \frac{1}{C - x}$ (8)

defines a solution of the differential equation

d y d x = y 2

$\frac{d y}{d x} = y^{2}$ (9)

on any interval of real numbers not containing the point x=C $x = C$ . Actually, Eq. (8) defines a one-parameter family of solutions of dy/dx=y2 $d y / d x = y^{2}$ , one for each value of the arbitrary constant or “parameter” C. With C=1 $C = 1$ we get the particular solution

y (x) = 1 1 - x

$y (x) = \frac{1}{1 - x}$

that satisfies the initial condition y(0)=1 $y (0) = 1$ . As indicated in Fig. 1.1.5, this solution is continuous on the interval (−∞,1) $(- \infty, 1)$ but has a vertical asymptote at x=1 $x = 1$ .

Example 8

Verify that the function y(x)=2x1/2−x1/2 ln x $y (x) = 2 x^{1/2} - x^{1/2} \ln x$ satisfies the differential equation

4 x 2 y'' + y = 0

$4 x^{2} y^{″} + y = 0$ (10)

for all x>0 $x > 0$ .

Solution

First we compute the derivatives

y' (x) = - 1 2 x - 1/2 ln x and y'' (x) = 1 4 x - 3/2 ln x - 1 2 x - 3/2 .

$\begin{array}{l} y^{′} (x) = - \frac{1}{2} x^{- 1/2} \ln x & and & y^{″} (x) = \frac{1}{4} x^{- 3/2} \ln x - \frac{1}{2} x^{- 3/2} \end{array} .$

Then substitution into Eq. (10) yields

4 x 2 y'' + y = 4 x 2 (1 4 x - 3/2 ln x - 1 2 x - 3/2) + 2 x 1/2 - x 1/2 ln x = 0

$4 x^{2} y ″ + y = 4 x^{2} (\frac{1}{4} x^{- 3/2} \ln x - \frac{1}{2} x^{- 3/2}) + 2 x^{1/2} - x^{1/2} \ln x = 0$

if x is positive, so the differential equation is satisfied for all x>0 $x > 0$ .

The fact that we can write a differential equation is not enough to guarantee that it has a solution. For example, it is clear that the differential equation

(y') 2 + y 2 = - 1

${(y^{′})}^{2} + y^{2} = - 1$ (11)

has no (real-valued) solution, because the sum of nonnegative numbers cannot be negative. For a variation on this theme, note that the equation

(y') 2 + y 2 = 0

${(y^{′})}^{2} + y^{2} = 0$ (12)

obviously has only the (real-valued) solution y(x)≡0 $y (x) \equiv 0$ . In our previous examples any differential equation having at least one solution indeed had infinitely many.

The order of a differential equation is the order of the highest derivative that appears in it. The differential equation of Example 8 is of second order, those in Examples 2 through 7 are first-order equations, and

y (4) + x 2 y (3) + x 5 y = sin x

$y^{(4)} + x^{2} y^{(3)} + x^{5} y = \sin x$

is a fourth-order equation. The most general form of an nth-order differential equation with independent variable x and unknown function or dependent variable y=y(x) $y = y (x)$ is

F (x, y, y', y'', \dots, y (n)) = 0,

$F (x, y, y^{′}, y^{″}, \dots, y^{(n)}) = 0,$ (13)

where F is a specific real-valued function of n+2 $n + 2$ variables.

Our use of the word solution has been until now somewhat informal. To be precise, we say that the continuous function u=u(x) $u = u (x)$ is a solution of the differential equation in (13) on the interval I provided that the derivatives u', u'', …, u(n) $u ′, u ″, \dots, u^{(n)}$ exist on I and

F (x, u, u', u'', \dots, u (n)) = 0

$F (x, u, u^{′}, u^{″}, \dots, u^{(n)}) = 0$

for all x in I. For the sake of brevity, we may say that u=u(x) $u = u (x)$ satisfies the differential equation in (13) on I.

Remark

Recall from elementary calculus that a differentiable function on an open interval is necessarily continuous there. This is why only a continuous function can qualify as a (differentiable) solution of a differential equation on an interval.

FIGURE 1.1.5.

The solution of y'=y2 $y ′ = y^{2}$ defined by y(x)=1/(1−x) $y (x) = 1 / (1 - x)$ .

Example 7

Continued Figure 1.1.5 shows the two “connected” branches of the graph y=1/(1−x) $y = 1 / (1 - x)$ . The left-hand branch is the graph of a (continuous) solution of the differential equation y'=y2 $y ′ = y^{2}$ that is defined on the interval (−∞, 1) $(- \infty, 1)$ . The right-hand branch is the graph of a different solution of the differential equation that is defined (and continuous) on the different interval (1, ∞) $(1, \infty)$ . So the single formula y(x)=1/(1−x) $y (x) = 1 / (1 - x)$ actually defines two different solutions (with different domains of definition) of the same differential equation y'=y2 $y ′ = y^{2}$ .

Example 9

If A and B are constants and

y (x) = A cos 3 x + B sin 3 x .

$y (x) = A \cos 3 x + B \sin 3 x .$ (14)

then two successive differentiations yield

y' (x) y'' (x) = = - 3 A sin 3 x + 3 B cos 3 x, - 9 A cos 3 x - 9 B sin 3 x = - 9 y (x)

$\begin{array}{rcl} y^{′} (x) & = & - 3 A \sin 3 x + 3 B \cos 3 x, \\ y^{″} (x) & = & - 9 A \cos 3 x - 9 B \sin 3 x = - 9 y (x) \end{array}$

for all x. Consequently, Eq. (14) defines what it is natural to call a two-parameter family of solutions of the second-order differential equation

y'' + 9 y = 0

$y^{″} + 9 y = 0$ (15)

on the whole real number line. Figure 1.1.6 shows the graphs of several such solutions.

Although the differential equations in (11) and (12) are exceptions to the general rule, we will see that an nth-order differential equation ordinarily has an n-parameter family of solutions—one involving n different arbitrary constants or parameters.

FIGURE 1.1.6.

The three solutions y1(x)=3 cos 3x, y2(x)=2 sin 3x $y_{1} (x) = 3 \cos 3 x, y_{2} (x) = 2 \sin 3 x$ , and y3(x)=−3 cos 3x+2 sin 3x $y_{3} (x) = - 3 \cos 3 x + 2 \sin 3 x$ of the differential equation y''+9y=0 $y ″ + 9 y = 0$ .

In both Eqs. (11) and (12), the appearance of y' $y ′$ as an implicitly defined function causes complications. For this reason, we will ordinarily assume that any differential equation under study can be solved explicitly for the highest derivative that appears; that is, that the equation can be written in the so-called normal form

y (n) = G (x, y, y', y'', \dots, y (n - 1)),

$y^{(n)} = G (x, y, y^{′}, y^{″}, \dots, y^{(n - 1)}),$ (16)

where G is a real-valued function of n+1 $n + 1$ variables. In addition, we will always seek only real-valued solutions unless we warn the reader otherwise.

All the differential equations we have mentioned so far are ordinary differential equations, meaning that the unknown function (dependent variable) depends on only a single independent variable. If the dependent variable is a function of two or more independent variables, then partial derivatives are likely to be involved; if they are, the equation is called a partial differential equation. For example, the temperature u=u(x,t) $u = u (x, t)$ of a long thin uniform rod at the point x at time t satisfies (under appropriate simple conditions) the partial differential equation

\partial u \partial t = k \partial 2 u \partial x 2,

$\frac{\partial u}{\partial t} = k \frac{\partial^{2} u}{\partial x^{2}},$

where k is a constant (called the thermal diffusivity of the rod). In Chapters 1 through 8 we will be concerned only with ordinary differential equations and will refer to them simply as differential equations.

In this chapter we concentrate on first-order differential equations of the form

d y d x = f (x, y) .

$\frac{d y}{d x} = f (x, y) .$ (17)

We also will sample the wide range of applications of such equations. A typical mathematical model of an applied situation will be an initial value problem, consisting of a differential equation of the form in (17) together with an initial condition y(x0)=y0 $y (x_{0}) = y_{0}$ . Note that we call y(x0)=y0 $y (x_{0}) = y_{0}$ an initial condition whether or not x0=0 $x_{0} = 0$ . To solve the initial value problem

d y d x = f (x, y), y (x 0) = y 0

$\begin{array}{l} \frac{d y}{d x} = f (x, y), & y (x_{0}) = y_{0} \end{array}$ (18)

means to find a differentiable function y=y(x) $y = y (x)$ that satisfies both conditions in Eq. (18) on some interval containing x0 $x_{0}$ .

Example 10

Given the solution y(x)=1/(C−x) $y (x) = 1 / (C - x)$ of the differential equation dy/dx=y2 $d y / d x = y^{2}$ discussed in Example 7, solve the initial value problem

d y d x = y 2, y (1) = 2 .

$\frac{d y}{d x} = y^{2}, y (1) = 2 .$

Solution

We need only find a value of C so that the solution y(x)=1/(C−x) $y (x) = 1 / (C - x)$ satisfies the initial condition y(1)=2 $y (1) = 2$ . Substitution of the values x=1 $x = 1$ and y=2 $y = 2$ in the given solution yields

2 = y (1) = 1 C - 1,

$2 = y (1) = \frac{1}{C - 1},$

so 2C−2=1 $2 C - 2 = 1$ , and hence C=32 $C = \frac{3}{2}$ . With this value of C we obtain the desired solution

y (x) = 1 3 2 - x = 2 3 - 2 x .

$y (x) = \frac{1}{\frac{3}{2} - x} = \frac{2}{3 - 2 x} .$

Figure 1.1.7 shows the two branches of the graph y=2/(3−2x) $y = 2 / (3 - 2 x)$ . The left-hand branch is the graph on (−∞,32) $(- \infty, \frac{3}{2})$ of the solution of the given initial value problem y'=y2, y(1)=2 $y ′ = y^{2}, y (1) = 2$ . The right-hand branch passes through the point (2,−2) $(2, - 2)$ and is therefore the graph on (32,∞) $(\frac{3}{2}, \infty)$ of the solution of the different initial value problem y'=y2, y(2)=−2 $y ′ = y^{2}, y (2) = - 2$ .

The central question of greatest immediate interest to us is this: If we are given a differential equation known to have a solution satisfying a given initial condition, how do we actually find or compute that solution? And, once found, what can we do with it? We will see that a relatively few simple techniques—separation of variables (Section 1.4), solution of linear equations (Section 1.5), elementary substitution methods (Section 1.6)—are enough to enable us to solve a variety of first-order equations having impressive applications.

FIGURE 1.1.7.

The solutions of y'=y2 $y ′ = y^{2}$ defined by y(x)=2/(3−2x) $y (x) = 2 / (3 - 2 x)$ .

1.1 Problems

In Problems 1 through 12, verify by substitution that each given function is a solution of the given differential equation. Throughout these problems, primes denote derivatives with respect to x.

y'=3x2 $y ′ = 3 x^{2}$ ; y=x3+7 $y = x^{3} + 7$
y'+2y=0 $y ′ + 2 y = 0$ ; y=3e−2x $y = 3 e^{- 2 x}$
y''+4y=0 $y ″ + 4 y = 0$ ; y1=cos 2x, y2=sin 2x $y_{1} = \cos 2 x, y_{2} = \sin 2 x$
y''=9y $y ″ = 9 y$ ; y1=e3x, y2=e−3x $y_{1} = e^{3 x}, y_{2} = e^{- 3 x}$
y'=y+2e−x $y ′ = y + 2 e^{- x}$ ; y=ex−e−x $y = e^{x} - e^{- x}$
y''+4y'+4y=0 $y ″ + 4 y ′ + 4 y = 0$ ; y1=e−2x, y2=xe−2x $y_{1} = e^{- 2 x}, y_{2} = x e^{- 2 x}$
y''−2y'+2y=0 $y ″ - 2 y ′ + 2 y = 0$ ; y1=ex cos x, y2=ex sin x $y_{1} = e^{x} \cos x, y_{2} = e^{x} \sin x$
y''+y=3 cos 2x, y1=cos x−cos 2x, y2=sin x−cos 2x $y ″ + y = 3 \cos 2 x, y_{1} = \cos x - \cos 2 x, y_{2} = \sin x - \cos 2 x$
y'+2xy2=0 $y ′ + 2 {x y}^{2} = 0$ ; y=11+x2 $y = \frac{1}{1 + x^{2}}$
x2y''+xy'−y=ln x $x^{2} y ″ + x y ′ - y = \ln x$ ; y1=x−ln x, y2=1x−ln x $y_{1} = x - \ln x, y_{2} = \frac{1}{x} - \ln x$
x2y''+5xy'+4y=0 $x^{2} y ″ + 5 x y ′ + 4 y = 0$ ; y1=1x2, y2=ln xx2 $y_{1} = \frac{1}{x^{2}}, y_{2} = \frac{\ln x}{x^{2}}$
x2y''−xy'+2y=0 $x^{2} y ″ - x y ′ + 2 y = 0$ ; y1=xcos(ln x), y2=xsin(ln x) $y_{1} = x \cos (\ln x), y_{2} = x \sin (\ln x)$

In Problems 13 through 16, substitute y=erx $y = e^{r x}$ into the given differential equation to determine all values of the constant r for which y=erx $y = e^{r x}$ is a solution of the equation.

3y'=2y $3 y ′ = 2 y$
4y''=y $4 y ″ = y$
y''+y'−2y=0 $y ″ + y ′ - 2 y = 0$
3y''+3y'−4y=0 $3 y ″ + 3 y ′ - 4 y = 0$

In Problems 17 through 26, first verify that y(x) satisfies the given differential equation. Then determine a value of the constant C so that y(x) satisfies the given initial condition. Use a computer or graphing calculator (if desired) to sketch several typical solutions of the given differential equation, and highlight the one that satisfies the given initial condition.

y'+y=0 $y ′ + y = 0$ ; y(x)=Ce−x, y(0)=2 $y (x) = {C e}^{- x}, y (0) = 2$
y'=2y $y ′ = 2 y$ ; y(x)=Ce2x, y(0)=3 $y (x) = C e^{2 x}, y (0) = 3$
y'=y+1 $y ′ = y + 1$ ; y(x)=Cex−1, y(0)=5 $y (x) = C e^{x} - 1, y (0) = 5$
y'=x−y $y ′ = x - y$ ; y(x)=Ce−x+x−1, y(0)=10 $y (x) = {C e}^{- x} + x - 1, y (0) = 10$
y'+3x2y=0 $y ′ + 3 x^{2} y = 0$ ; y(x)=Ce−x3, y(0)=7 $y (x) = {C e}^{- x^{3}}, y (0) = 7$
eyy'=1 $e^{y} y ′ = 1$ ; y(x)=ln (x+C), y(0)=0 $y (x) = \ln (x + C), y (0) = 0$
xdydx+3y=2x5 $x \frac{d y}{d x} + 3 y = 2 x^{5}$ ; y(x)=14x5+Cx−3, y(2)=1 $y (x) = \frac{1}{4} x^{5} + C x^{- 3}, y (2) = 1$
xy'−3y=x3 $x y ′ - 3 y = x^{3}$ ; y(x)=x3(C+ln x), y(1)=17 $y (x) = x^{3} (C + \ln x), y (1) = 17$
y'=3x2(y2+1) $y ′ = 3 x^{2} (y^{2} + 1)$ ; y(x)=tan(x3+C), y(0)=1 $y (x) = \tan (x^{3} + C), y (0) = 1$
y'+ytanx=cos x $y ′ + y \tan x = \cos x$ ; y(x)=(x+C) cos x, y(π)=0 $y (x) = (x + C) \cos x, y (π) = 0$

In Problems 27 through 31, a function y=g(x) $y = g (x)$ is described by some geometric property of its graph. Write a differential equation of the form dy/dx=f(x,y) $d y / d x = f (x, y)$ having the function g as its solution (or as one of its solutions).

The slope of the graph of g at the point (x, y) is the sum of x and y.
The line tangent to the graph of g at the point (x, y) intersects the x-axis at the point (x/2,0) $(x / 2, 0)$ .
Every straight line normal to the graph of g passes through the point (0, 1). Can you guess what the graph of such a function g might look like?
The graph of g is normal to every curve of the form y=x2+k $y = x^{2} + k$ (k is a constant) where they meet.
The line tangent to the graph of g at (x, y) passes through the point (−y,x) $(- y, x)$ .

Differential Equations as Models

In Problems 32 through 36, write—in the manner of Eqs. (3) through (6) of this section—a differential equation that is a mathematical model of the situation described.

The time rate of change of a population P is proportional to the square root of P.
The time rate of change of the velocity v of a coasting motorboat is proportional to the square of v.
The acceleration dv/dt $d v / d t$ of a Lamborghini is proportional to the difference between 250 km/h and the velocity of the car.
In a city having a fixed population of P persons, the time rate of change of the number N of those persons who have heard a certain rumor is proportional to the number of those who have not yet heard the rumor.
In a city with a fixed population of P persons, the time rate of change of the number N of those persons infected with a certain contagious disease is proportional to the product of the number who have the disease and the number who do not.

In Problems 37 through 42, determine by inspection at least one solution of the given differential equation. That is, use your knowledge of derivatives to make an intelligent guess. Then test your hypothesis.

y''=0 $y ″ = 0$
y'=y $y ′ = y$
xy'+y=3x2 $x y ′ + y = 3 x^{2}$
(y')2+y2=1 ${(y ′)}^{2} + y^{2} = 1$
y'+y=ex $y ′ + y = e^{x}$
y''+y=0 $y ″ + y = 0$

Problems 43 through 46 concern the differential equation

d x d t = k x 2,

$\frac{d x}{d t} = {k x}^{2},$

where k is a constant.

1. If k is a constant, show that a general (one-parameter) solution of the differential equation is given by x(t)=1/(C−kt) $x (t) = 1 / (C - k t)$ , where C is an arbitrary constant.
2. Determine by inspection a solution of the initial value problem x'=kx2, x(0)=0 $x ′ = {k x}^{2}, x (0) = 0$ .
1. Assume that k is positive, and then sketch graphs of solutions of x'=kx2 $x ′ = {k x}^{2}$ with several typical positive values of x(0) $x (0)$ .
2. How would these solutions differ if the constant k were negative?
Suppose a population P of rodents satisfies the differential equation dP/dt=kP2 $d P / d t = {k P}^{2}$ . Initially, there are P(0)=2 $P (0) = 2$ rodents, and their number is increasing at the rate of dP/dt=1 $d P / d t = 1$ rodent per month when there are P=10 $P = 10$ rodents. Based on the result of Problem 43, how long will it take for this population to grow to a hundred rodents? To a thousand? What’s happening here?

FIGURE 1.1.8.

Graphs of solutions of the equation dy/dx=y2. $d y / d x = y^{2} .$
Suppose the velocity v of a motorboat coasting in water satisfies the differential equation dv/dt=kv2 $d v / d t = {k v}^{2}$ . The initial speed of the motorboat is v(0)=10 $v (0) = 10$ meters per second (m/s), and v is decreasing at the rate of 1 m/s2 ${m/s}^{2}$ when v=5 m/s $v = 5 m/s$ . Based on the result of Problem 43, long does it take for the velocity of the boat to decrease to 1 m/s? To 110 m/s $\frac{1}{10} m/s$ ? When does the boat come to a stop?
In Example 7 we saw that y(x)=1/(C−x) $y (x) = 1 / (C - x)$ defines a one-parameter family of solutions of the differential equation dy/dx=y2 $d y / d x = y^{2}$ .
1. Determine a value of C so that y(10)=10 $y (10) = 10$ .
2. Is there a value of C such that y(0)=0 $y (0) = 0$ ? Can you nevertheless find by inspection a solution of dy/dx=y2 $d y / d x = y^{2}$ such that y(0)=0 $y (0) = 0$ ?
3. Figure 1.1.8 shows typical graphs of solutions of the form y(x)=1/(C−x) $y (x) = 1 / (C - x)$ . Does it appear that these solution curves fill the entire xy-plane? Can you conclude that, given any point (a, b) in the plane, the differential equation dy/dx=y2 $d y / d x = y^{2}$ has exactly one solution y(x) satisfying the condition y(a)=b $y (a) = b$ ?
1. Show that y(x)=Cx4 $y (x) = C x^{4}$ defines a one-parameter family of differentiable solutions of the differential equation xy'=4y $x y ′ = 4 y$ (Fig. 1.1.9).
2. Show that
  
  $y (x) = {- x 4 x 4 if x < 0, if x ≧ 0$ $y (x) = {\begin{aligned} - x^{4} & if x < 0, \\ x^{4} & if x ≧ 0 \end{aligned}$
  
  defines a differentiable solution of xy'=4y $x y ′ = 4 y$ for all x, but is not of the form y(x)=Cx4 $y (x) = C x^{4}$ .
3. Given any two real numbers a and b, explain why—in contrast to the situation in part (c) of Problem 47—there exist infinitely many differentiable solutions of xy'=4y $x y ′ = 4 y$ that all satisfy the condition y(a)=b $y (a) = b$ .
FIGURE 1.1.9.

The graph y=Cx4 $y = C x^{4}$ for various values of C.

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Table of Contents for 1 First-Order Differential Equations

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Table of Contents for
1 First-Order Differential Equations