2.2 Separable Equations

INTRODUCTION

Consider the first-order equations dy/dx = f(x, y). When f does not depend on the variable y, that is, f(x, y) = g(x), the differential equation

(1)

can be solved by integration. If g(x) is a continuous function, then integrating both sides of (1) gives the solution y = ∫ g(x) dx = G(x) + c, where G(x) is an antiderivative (indefinite integral) of g(x). For example, if dy/dx = 1 + e2x, then y = ∫ (1 + e2x) dx or defined on (−, ).

A Definition

Equation (1), as well as its method of solution, is just a special case when f in dy/dx = f(x, y) is a product of a function of x and a function of y.

DEFINITION 2.2.1 Separable Equation

A first-order differential equation of the form

is said to be separable or to have separable variables.

For example, the differential equations

are separable and nonseparable, respectively. To see this, note that in the first equation we can factor as

but in the second equation there is no way of writing y + cos x as a product of a function of x times a function of y.

Method of Solution

Observe that by dividing by the function h(y), a separable equation can be written as

, (2)

where, for convenience, we have denoted 1/h(y) by p(y). From this last form we can see immediately that (2) reduces to (1) when h(y) = 1.

Now if y = ϕ(x) represents a solution of (2), we must have p(ϕ(x))ϕ′(x) = g(x), and therefore,

(3)

But dy = ϕ′(x) dx, and so (3) is the same as

(4)

where H(y) and G(x) are antiderivatives of p(y) = 1/h(y) and g(x), respectively.

Equation (4) indicates the procedure for solving separable equations. A one-parameter family of solutions, usually given implicitly, is obtained by integrating both sides of the differential form p(y) dy = g(x) dx.

In solving first-order DEs, use only one constant.

There is no need to use two constants in the integration of a separable equation, because if we write H(y) + c1 = G(x) + c2, then the difference c2 − c1 can be replaced by a single constant c, as in (4). In many instances throughout the chapters that follow, we will relabel constants in a manner convenient to a given equation. For example, multiples of constants or combinations of constants can sometimes be replaced by a single constant.

EXAMPLE 1 Solving a Separable DE

Solve (1 + x) dy − y dx = 0.

SOLUTION

Dividing by (1 + x)y, we can write dy/y = dx/(1 + x), from which it follows that

and so

Relabeling ± by c then gives defined on (−, ). ≡

In the solution of Example 1, because each integral results in a logarithm, a judicious choice for the constant of integration is ln |c| rather than c. Rewriting the second line of the solution as ln |y| = ln |1 + x| + ln |c| enables us to combine the terms on the right-hand side by the properties of logarithms. From ln |y| = ln |c(1 + x)|, we immediately get y = c(1 + x). Even if the indefinite integrals are not all logarithms, it may still be advantageous to use ln |c|. However, no firm rule can be given.

In Section 1.1 we have already seen that a solution curve may be only a segment or an arc of the graph of an implicit solution G(x, y) = 0.

EXAMPLE 2 Solution Curve

Solve the initial-value problem , .

SOLUTION

By rewriting the equation as y dy = −x dx we get

    and    

We can write the result of the integration as x2 + y2 = c2 by replacing the constant 2c1 by c2. This solution of the differential equation represents a one-parameter family of concentric circles centered at the origin.

Now when x = 4, y = −3, so that 16 + 9 = 25 = c2. Thus the initial-value problem determines the circle with radius 5. Because of its simplicity, we can solve this implicit solution for an explicit solution that satisfies the initial condition. We have seen this solution as y = ϕ2(x) or , in Example 8 of Section 1.1. A solution curve is the graph of a differentiable function. In this case the solution curve is the lower semicircle, shown in blue in FIGURE 2.2.1, that contains the point (4, −3). ≡

A graph consists of 4 concentric circles centered at the origin. The radii of the circles are as follows: 1, 3, 5, and 6. On the circle with radius 5, the lower semicircle is highlighted, and point (4, negative 3) is marked.

FIGURE 2.2.1 Solution curve for IVP in Example 2

EXAMPLE 3 Initial-Value Problem

Solve the initial-value problem .

SOLUTION

Rewriting the equation as we see that

Now y(0) = 0 in the last equation gives tan−1 0 = 0 + c and so c = 0. The solution is tan−1 y = x or with interval of definition . ≡

Most of us would write the solution tan−1 y = x + c in Example 3 as y = tan(x + c). But we have to be a bit careful here. If we simply specify a value of c, then y = tan(x + c) actually defines an infinite number of particular solutions corresponding to an infinite number of intervals of definition. For example, for c = 0

, ,

are solutions of the differential equation. See FIGURE 2.2.2. Only the red graph is the solution of the IVP in Example 3.

A graph has 3 curves. 4 vertical dashed lines are marked passing through the following values of x: negative three pi over two, negative pi over two, pi over two, and three pi over two, respectively. The curves are translations by pi over 2 of the curve labeled y equals tangent of x. It enters the third quadrant to the right of x equals negative pi over 2, goes up and to the right with decreasing steepness through the origin, then goes up and to the right with increasing steepness as it nears x equals pi over 2.

FIGURE 2.2.2 Solutions of DE in Example 3

Losing a Solution

Some care should be exercised when separating variables, since the variable divisors could be zero at a point. Specifically, if r is a zero of the function h(y), then substituting y = r into dy/dx = g(x)h(y) makes both sides zero; in other words, y = r is a constant solution of the differential equation. But after separating variables, observe that the left side of dy/h(y) = g(x)dx is undefined at r. As a consequence, y = r may not show up in the family of solutions obtained after integration and simplification. Recall, such a solution is called a singular solution.

EXAMPLE 4 Losing a Solution

Solve = y2 − 4.

SOLUTION

We put the equation in the form

or . (5)

The second equation in (5) is the result of using partial fractions on the left side of the first equation. Integrating and using the laws of logarithms gives

Here we have replaced 4c1 by c2. Finally, after replacing by c and solving the last equation for y, we get the one-parameter family of solutions

. (6)

Now if we factor the right side of the differential equation as dy/dx = (y − 2)(y + 2), we know from the discussion in Section 2.1 that y = 2 and y = −2 are two constant (equilibrium) solutions. The solution y = 2 is a member of the family of solutions defined by (6) corresponding to the value c = 0. However, y = −2 is a singular solution; it cannot be obtained from (6) for any choice of the parameter c. This latter solution was lost early on in the solution process. Inspection of (5) clearly indicates that we must preclude y = ±2 in these steps. ≡

EXAMPLE 5 An Initial-Value Problem

Solve the initial-value problem

, .

SOLUTION

Dividing the equation by ey cos x gives

Before integrating, we use termwise division on the left side and the trigonometric identity sin 2x = 2 sin x cos x on the right side. Then

(7)

yields

The initial condition y = 0 when x = 0 implies c = 4. Thus a solution of the initial-value problem is

(8) ≡

Use of Computers

In the Remarks at the end of Section 1.1 we mentioned that it may be difficult to use an implicit solution G(x, y) = 0 to find an explicit solution y = ϕ(x). Equation (8) shows that the task of solving for y in terms of x may present more problems than just the drudgery of symbol pushing—it simply can’t be done! Implicit solutions such as (8) are somewhat frustrating; neither the graph of the equation nor an interval over which a solution satisfying y(0) = 0 is defined is apparent. The problem of “seeing” what an implicit solution looks like can be overcome in some cases by means of technology. One way* of proceeding is to use the contour plot application of a CAS. Recall from multivariate calculus that for a function of two variables z = G(x, y) the two-dimensional curves defined by G(x, y) = c, where c is constant, are called the level curves of the function. With the aid of a CAS we have illustrated in FIGURE 2.2.3 some of the level curves of the function G(x, y) = ey + ye–y + e–y + 2 cos x. The family of solutions defined by (7) are the level curves G(x, y) = c. FIGURE 2.2.4 illustrates, in blue, the level curve G(x, y) = 4, which is the particular solution (8). The red curve in Figure 2.2.4 is the level curve G(x, y) = 2, which is the member of the family G(x, y) = c that satisfies y(π/2) = 0.

A family of level curves symmetrical about the vertical axis is graphed. The curves at the top become more flat as y increases. They enter the second quadrant, go down and to the right to a low point on the vertical axis, then go up and to the right into the first quadrant. 4 curves in the middle of the range go down and to the right with increasing steepness, then with decreasing steepness to a low point on the vertical axis, go up and to the right with increasing steepness, then with decreasing steepness. The curves below become more flat as y decreases. They enter the third quadrant, go down and to the right to a low point on the vertical axis, then go up and to the right into the fourth quadrant.

FIGURE 2.2.3 Level curves G(x, y) = c, where G(x, y) = ey + ye−y + e−y + 2 cos x

The graph consists of two curves. The first curve labeled c equals two enters the second quadrant, goes down and to the right, passing through the origin. The curve then goes up and to the right into the first quadrant. The second curve labeled c equals four enters the second quadrant, goes down and to the right, and intersects the negative horizontal axis. It continues into the third quadrant, goes down and to the right, and intersects the negative vertical axis. The curve continues into the fourth quadrant, goes up and to the right, passing through (pi over 2, zero), then into the first quadrant.

FIGURE 2.2.4 Level curves c = 2 and c = 4

If an initial condition leads to a particular solution by finding a specific value of the parameter c in a family of solutions for a first-order differential equation, it is a natural inclination for most students (and instructors) to relax and be content. However, a solution of an initial-value problem may not be unique. We saw in Example 4 of Section 1.2 that the initial-value problem

,     (9)

has at least two solutions, y = 0 and y = x4. We are now in a position to solve the equation.

Separating variables and integrating gives . Solving for y and replacing by the symbol c yields

. (10)

Each of the functions in the family given in (10) is a solution of equation (9) on the interval (−, ) provided we take c ≥ 0. See Problem 56 in Exercises 2.2. Now when we substitute x = 0, y = 0 in (10) we see that c = 0. Therefore is a solution of the IVP. But note that the solution y = 0 is not a member of the family of solutions (10). This singular solution was lost in the solution process by dividing by . The initial-value problem (9) actually possesses many more solutions, since for any choice of the parameter a ≥ 0 the piecewise-defined function

satisfies both the differential equation and the initial condition. See FIGURE 2.2.5.

The graph consists of a solid line representing a equals zero and two dashed lines representing a greater than zero. The solid curve begins at the origin and goes up and right with increasing steepness. The first dashed curve begins at the origin and goes right. It then goes up and right with increasing steepness, parallel to curve 1. The third curve begins at the origin, goes more to the right than curve 2. It then goes up and to the right with increasing steepness, and is parallel to the first 2 curves.

FIGURE 2.2.5 Piecewise-defined solutions of (9)

An Integral-Defined Function

In (ii) of the Remarks at the end of Section 1.1 it was pointed out that a solution method for a certain kind of differential equation may lead to an integral-defined function. This is especially true for separable differential equations because integration is the method of solution. For example, if g is continuous on some interval I containing x0 and x, then a solution of the simple initial-value problem dy/dx = g(x), y(x0) = y0 defined on I is given by

.

To see this, we have immediately from (12) of Section 1.1 that dy/dx = g(x) and y(x0) = y0 because . When is nonelementary, that is, cannot be expressed in terms of elementary functions, the form may be the best we can do in obtaining an explicit solution of an IVP. The next example illustrates this idea.

EXAMPLE 6 An Initial-Value Problem

Solve .

SOLUTION

The function is continuous on the interval but its antiderivative is not an elementary function. Using t as a dummy variable of integration, we integrate both sides of the given differential equation:

Using the initial condition y(2) = 6 we obtain the solution . ≡

The procedure illustrated in Example 6 works equally well on separable equations dy/dx = g(x)f(y) where, say, f(y) possesses an elementary antiderivative but g(x) does not possess an elementary antiderivative. See Problems 31–34 in Exercises 2.2.

REMARKS

In some of the preceding examples we saw that the constant in the one-parameter family of solutions for a first-order differential equation can be relabeled when convenient. Also, it can easily happen that two individuals solving the same equation correctly arrive at dissimilar expressions for their answers. For example, by separation of variables, we can show that one-parameter families of solutions for the DE (1 + y2) dx + (1 + x2) dy = 0 are

    or    

As you work your way through the next several sections, keep in mind that families of solutions may be equivalent in the sense that one family may be obtained from another by either relabeling the constant or applying algebra and trigonometry. See Problems 27 and 28 in Exercises 2.2.

2.2 Exercises Answers to selected odd-numbered problems begin on page ANS-2.

In Problems 1–22, solve the given differential equation by separation of variables.

In Problems 23–30, find an explicit solution of the given initial-value problem.

  1. , x(π/4) = 1
  2. , y(2) = 2
  3. , y(−1) = −1
  4. , y(0) =
  5. , y(0) = /2
  6. (1 + x4) dy + x(1 + 4y2) dx = 0, y(1) = 0

In Problems 31–34, proceed as in Example 6 and find an explicit solution of the given initial-value problem.

In Problems 35–38, find an explicit solution of the given initial-value problem. Determine the exact interval I of definition of each solution by analytical methods. Use a graphing utility to plot the graph of each solution.

    1. Find a solution of the initial-value problem consisting of the differential equation in Example 4 and the initial conditions y(0) = 2, y(0) = −2, y() = 1.
    2. Find the solution of the differential equation in Example 4 when ln c1 is used as the constant of integration on the left-hand side in the solution and 4 ln c1 is replaced by ln c. Then solve the same initial-value problems in part (a).
  6. Find a solution of that passes through the indicated points.

    (a) (0, 1)

    (b) (0, 0)

    (c) (, )

    (d) (2, )

  7. Find a singular solution of Problem 21. Of Problem 22.
  8. Show that an implicit solution of

    2x sin2 y dx − (x2 + 10) cos y dy = 0

    is given by ln(x2 + 10) csc y = c. Find the constant solutions, if any, that were lost in the solution of the differential equation.

Often a radical change in the form of the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. In Problems 43–46, find an explicit solution of the given initial-value problem. Use a graphing utility to plot the graph of each solution. Compare each solution curve in a neighborhood of (0, 1).

  1. ,     
  2. ,     
  3. ,     
  4. ,     
  5. Every autonomous first-order equation dy/dx = f(y) is separable. Find explicit solutions y1(x), y2(x), y3(x), and y4(x) of the differential equation dy/dx = y − y3 that satisfy, in turn, the initial conditions y1(0) = 2, y2(0) = , y3(0) = −, and y4(0) = −2. Use a graphing utility to plot the graphs of each solution. Compare these graphs with those predicted in Problem 19 of Exercises 2.1. Give the exact interval of definition for each solution.
    1. The autonomous first-order differential equation dy/dx = 1/(y − 3) has no critical points. Nevertheless, place 3 on a phase line and obtain a phase portrait of the equation. Compute d2y/dx2 to determine where solution curves are concave up and where they are concave down (see Problems 35 and 36 in Exercises 2.1). Use the phase portrait and concavity to sketch, by hand, some typical solution curves.
    2. Find explicit solutions y1(x), y2(x), y3(x), and y4(x) of the differential equation in part (a) that satisfy, in turn, the initial conditions y1(0) = 4, y2(0) = 2, y3(1) = 2, and y4(−1) = 4. Graph each solution and compare with your sketches in part (a). Give the exact interval of definition for each solution.

In Problems 49–54, use a technique of integration or a substitution to find an explicit solution of the given differential equation or initial-value problem.

Discussion Problems

    1. Explain why the interval of definition of the explicit solution y = ϕ2(x) of the initial-value problem in Example 2 is the open interval (−5, 5).
    2. Can any solution of the differential equation cross the x-axis? Do you think that x2 + y2 = 1 is an implicit solution of the initial-value problem dy/dx = −x/y, y(1) = 0?
  2. On page 48 we showed that a one-parameter family of solutions of the first-order differential equation is for c ≥ 0. Each solution in this family is defined on the interval (−, ). The last statement is not true if we choose c to be negative. For c = −1, explain why y = ( is not a solution of the DE on (−, ). Find an interval of definition I on which y = ( is a solution of the DE.
  3. In Problems 47 and 48 we saw that every autonomous first-order differential equation dy/dx = f(y) is separable. Does this fact help in the solution of the initial-value problem y(0) = ? Discuss. Sketch, by hand, a plausible solution curve of the problem.

  4. (a) Solve the two initial-value problems:

    (b) Show that there are more than 1.65 million digits in the y-coordinate of the point of intersection of the two solution curves in part (a).

  5. Find a function whose square plus the square of its derivative is 1.

  6. (a) The differential equation in Problem 27 is equivalent to the normal form

    in the square region in the xy-plane defined by |x| < 1, |y| < 1. But the quantity under the radical is nonnegative also in the regions defined by |x| > 1, |y| > 1. Sketch all regions in the xy-plane for which this differential equation possesses real solutions.

    (b) Solve the DE in part (a) in the regions defined by |x| > 1, |y| > 1. Then find an implicit and an explicit solution of the differential equation subject to y(2) = 2.

Mathematical Models

  1. Hawking Radiation      According to the English theoretical physicist Stephen Hawking (1942–2018), the mass m(t) of a black hole decreases over time due to radiation. A model for the mass is given by

    Solve for m(t) when m(0) = m0.

  2. Hanging Chain      Suppose a flexible steel chain is hanging under its own weight and its ends are attached at the same height. See the photo below.

    (a) When determining the shape of the chain, one has to first solve the differential equation

    where a is a constant. Solve for u(x) when u(0) = 0.

    (b) The function u(x) found in part (a) is related to the shape of the chain y(x) by the differential equation

    Solve for y(x) when y(0) = 1/a. Identify the shape of the hanging chain.

    Thick steel chains hang between posts to form a barrier.

    © Stoyan G. Ivanov/Shutterstock

    Hanging chain in Problem 62

  3. Tautochrone Problem      The tautochrone problem consists of finding a curve C such that the time of descent of a bead sliding without friction on C to its lowest point is independent of its starting point. If the curve C in FIGURE 2.2.6 is a tautochrone, then the three beads starting from different positions corresponding to x-coordinates x1, x2, and x3 will reach the origin (0, 0) in the same time. The Dutch mathematician and astronomer Christiaan Huygens (1629–1695) was the first to publish the solution of the tautochrone problem in 1673.

    In determining such a curve, the following differential equation must be solved:

    where a is a constant. Find parametric equations for C by solving the DE in the form

    Use the substitution y = a sin2 ϕ.

    A curve labeled C is graphed. It begins at the origin and goes up and to the right with increasing steepness. Three points are marked on the curve, corresponding to x subscript one, x subscript two, and x subscript three on the horizontal axis. Arrows tangent to the curve are marked, pointing down and to the left at each of the three points.

    FIGURE 2.2.6 Tautochrone in Problem 63

  4. Chemical Reaction      In a certain second-order chemical reaction involving a single reactant, the rate of reaction is given by

    where k and A are constants and x(t) is the number of grams of the new chemical formed in time t. Solve for x(t) when x(0) = 0.

  5. Suspension Bridge      In (16) of Section 1.3 we saw that a mathematical model for the shape of a flexible cable strung between two vertical supports is

    (11)

    where W denotes the portion of the total vertical load between the points P1 and P2 shown in Figure 1.3.9. The DE (11) is separable under the following conditions that describe a suspension bridge.

    Let us assume that the x- and y-axes are as shown in FIGURE 2.2.7—that is, the x-axis runs along the horizontal roadbed, and the y-axis passes through (0, a), which is the lowest point on one cable over the span of the bridge, coinciding with the interval [−L/2, L/2]. In the case of a suspension bridge, the usual assumption is that the vertical load in (11) is only a uniform roadbed distributed along the horizontal axis. In other words, it is assumed that the weight of all cables is negligible in comparison to the weight of the roadbed and that the weight per unit length of the roadbed (say, pounds per horizontal foot) is a constant r. Use this information to set up and solve an appropriate initial-value problem from which the shape (a curve with equation y = (x)) of each of the two cables in a suspension bridge is determined. Express your solution of the IVP in terms of the sag h and span L shown in Figure 2.2.7.

    A suspension bridge is graphed. The curve representing the cable begins in the second quadrant, goes down and to the right to (a, 0), then goes up and to the right into the first quadrant. The roadbed or load is along the horizontal axis. The height between the lowest point of the curve and the top of the bridge is h sag. The span L of the bridge is divided by the vertical axis passing through (a, 0).

    FIGURE 2.2.7 Shape of a cable in Problem 65

Computer Lab Assignments

    1. Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation Experiment with different numbers of level curves as well as various rectangular regions defined by a ≤ x ≤ b, c ≤ y ≤ d.
    2. On separate coordinate axes plot the graphs of the particular solutions corresponding to the initial conditions: y(0) = −1; y(0) = 2; y(−1) = 4; y(−1) = −3.
    1. Find an implicit solution of the IVP

      ,

    2. Use part (a) to find an explicit solution y = (x) of the IVP.
    3. Consider your answer to part (b) as a function only. Use a graphing utility or a CAS to graph this function, and then use the graph to estimate its domain.
    4. With the aid of a root-finding application of a CAS, determine the approximate largest interval I of definition of the solution y = (x) in part (b). Use a graphing utility or a CAS to graph the solution curve for the IVP on this interval.
    1. Use a CAS and the concept of level curves to plot representative graphs of members of the family of solutions of the differential equation Experiment with different numbers of level curves as well as various rectangular regions in the xy-plane until your result resembles FIGURE 2.2.8.
    2. On separate coordinate axes, plot the graph of the implicit solution corresponding to the initial condition y(0) = . Use a colored pencil to mark off that segment of the graph that corresponds to the solution curve of a solution that satisfies the initial condition. With the aid of a root-finding application of a CAS, determine the approximate largest interval I of definition of the solution . [Hint: First find the points on the curve in part (a) where the tangent is vertical.]
    3. Repeat part (b) for the initial condition y(0) = −2.
      A family of level curves is graphed. 2 ellipses are marked around (0, 0) and (1, 2). One set of curves enters the second quadrant, goes down and to the right, then goes right around the ellipse at (1, 2), goes down, then goes down and to the right into the fourth quadrant. The other set of curves enters the second quadrant, go down and to the right, then down around the ellipse at (0, 0), then right and down and to the right into the fourth quadrant.

      FIGURE 2.2.8 Level curves in Problem 68

 

*In Section 2.6 we discuss several other ways of proceeding that are based on the concept of a numerical solver.