Section 5.3 Separable Differential Equations D.3 Separable Differential Equations Sketching solutions of a differential equation using its direction field is a powerful technique, and it provides a wealth of information about the solutions. However, valuable as they are, direction fields do not produce the actual solutions of a differential equation. In this section, we examine methods that lead to the solutions of certain differential equations in terms of an algebraic expression (often called an analytical solution). The equations we consider are first order and belong to a class called separable equations. Method of Solution The most general first-order differential equation has the form y' (t) = f (t, y), where f (t, y) is an expression that may involve both the independent variable t and the unknown function y. We have a chance of solving such an equation if it can be written in the form g(y) y' (t) = h(t). In the equation g(y) y' (t) = h(t), the factor g(y) involves only y, and h(t) involves only t; that is, the variables have been separated. An equation that can be written in this form is said to be separable. Note If the equation has the form y ' (t) = f (t) (that is, the right side depends only on t), then solving the equation amounts to finding the antiderivatives of f. In general, we solve a separable differential equation by integrating both sides of the equation with respect to t: g(y) y' (t) d t = h(t) dt Integrate both sides with respect to t. d y g(y) d y = h(t) dt. Change variables on left; d y = y' (t) dt. The fact that d y = y' (t) d t on the left side of the equation leaves us with two integrals to evaluate, one with respect to y and one with respect to t. Finding a solution depends on evaluating these integrals. QUICK CHECK Which of the following equations are separable? (A) y' (t) = y + t, (B) y' t y (t) =, and (C) t + y' (x) = e x+y EXAMLE A separable equation Find a function that satisfies the following initial value problem. y' (t) = y e -t, y(0) =, for t 0. SOLUTION The equation is written in separable form by dividing both sides of the equation by y to give both sides of the equation with respect to t and evaluate the resulting integrals: Note y' (t) = e -t. We now integrate y Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Chapter 5 Differential Equations In practice, the change of variable on the left side is often omitted, and we go directly to the second step, which is to integrate the left side with respect to y and the right side with respect to t. y y' (t) d t = e -t d t d y = y e -t d t Change variables on left side. - = -e -t + C. Evaluate integrals. y Note Notice that each integration produces a constant of integration. The two constants of integration may be combined into one. Solving for y gives the general solution The initial condition y(0) = implies that y(t) =. e -t - C y(0) = = =. e 0 - C - C It follows that C = -, so the solution to the initial value problem is y(t) =. e -t + The solution (Figure D.6) passes through 0, and increases to approach the asymptote y = because lim =. t e -t + Figure D.6 Related Exercises 5 6 Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 3 QUICK CHECK Write y' (t) = t + in separated form. y 3 EXAMLE Another separable equation Find the solutions of the equation y' (x) = e -y sin x subject to the three different initial conditions y(0) =, y π =, and y(0) = -3. SOLUTION Writing the equation in the form e y y' (x) = sin x, we see that it is separable. Integrating both sides with respect to x, we have e y y' (x) d x = sin x d x e y d y = sin x d x Change variables on left side. e y = -cos x + C. Evaluate integrals. The general solution y is found by taking logarithms of both sides of this equation: y = ln (C - cos x). The three initial conditions are now used to evaluate the constant C for the three solutions: y (0) = = ln(c - cos 0) = ln(c - ) e = C - C = e +, y π = = ln C - cos π = ln C C = e/, and y(0) = -3-3 = ln(c - cos 0) = ln(c - ) e -3 = C - C = e -3 +. Substituting these values of C into the general solution gives the solutions of the three initial value problems (Figure D.7). The points indicate the initial condition for each solution. Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
4 Chapter 5 Differential Equations C C = e + y C =.6487 C = e / C = e -3 + (0, ) π, - 4 6 8 0 x - - -3 (0, -3) Figure D.7 QUICK CHECK 3 Find the value of the constant C in Example with the initial condition y(π) = 0. Related Exercises 5 6 Even if we can evaluate the integrals necessary to solve a separable equation, the solution may not be easily expressed in an explicit form. Here is an example of a solution that is best left in implicit form. EXAMLE 3 An implicit solution Find and graph the solution of the initial value problem SOLUTION cos y y' (t) = sin t cos t, y(0) = The equation is already in separated form. Integrating both sides with respect to t, we have Note For the integral on the right side, we use the substitution u = sin t. The integral becomes u du = u 3 + C. 3 π 6. cos y d y = sin t cos t d t Integrate both sides. sin y = 3 sin3 t + C. Evaluate integrals. When imposing the initial condition in this case, it is best to leave the general solution in implicit form. Substituting t = 0 and π y = into the general solution, we find that 6 Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 5 Therefore, the solution of the initial value problem is sin π = sin 3 0 + C or C = 6 3 sin y = sin 3 t +. 3 In order to graph the solution in this implicit form, it is easiest to use graphing software. The result is shown in Figure D.8.. initial condition y(0) = 5 π 6 C = y 4 y(0) = π 6 0, 5 π 6 y(0) = - 7 π 6 0, π 6-4 - 4 x - -4 0, - 7 π 6 Figure D.8 Note Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
6 Chapter 5 Differential Equations Care must be used when graphing and interpreting implicit solutions. The graph of sin y = sin 3 t + 3 is a family of an infinite number of curves. y 3-4 -3 - - 3 - - -3-4 x You must choose the curve that satisfies the initial condition, as shown in Figure D.8. Related Exercises 7 3 QUICK CHECK 4 Find the value of the constant C in Example 3 with the initial condition y π 6 = 0. Logistic Equation Revisited In Section D., we introduced the logistic equation, which is commonly used for modeling populations, epidemics, and the spread of rumors. In Section D., we investigated the direction field associated with the logistic equation. It turns out that the logistic equation is a separable equation, so we now have the tools needed to solve it. Note The derivation of the logistic equation is discussed in Section D.5. EXAMLE 4 Logistic population growth Assume 50 fruit flies are in a large jar at the beginning of an experiment. Let (t) be the number of fruit flies in the jar t days later. At first, the population grows exponentially, but due to limited space and food supply, the growth rate decreases and the population is prevented from growing without bound. This experiment is modeled by the logistic equation d = 0. - d t 300, for t 0, together with the initial condition (0) = 50. Solve this initial value problem. SOLUTION We see that the equation is separable by writing it in the form - d = 0.. 300 dt Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 7 Integrating both sides with respect to t leads to the equation d - = 0. d t. () 300 The integral on the right side of equation () is 0. d t = 0. t + C. Because the integrand on the left side is a rational function in, we use partial fractions. You should verify that and therefore, - = 300 300 = + (300 - ) 300 - - d = 300 + 300 - d = ln 300 - + C. After integration, equation () becomes ln 300 - = 0. t + C. () The next step is to solve for, which is tangled up inside the logarithm. We first exponentiate both sides of equation () to obtain 300 - = e C e 0. t. We can remove the absolute value on the left side of equation () by writing = ±e C e 0. t. 300 - At this point, a useful trick simplifies matters. Because C is an arbitrary constant, ±e C is also an arbitrary constant, we rename ±e C as C. We now have Note = C e 0. t. (3) 300 - There are not many times in mathematics when we can redefine a constant in the middle of a calculation. When working with arbitrary constants, it may be possible, if it is done carefully. Solving equation (3) for and replacing C by C gives the general solution (t) = 300 + C e -0. t. Figure D.9 shows the general solution, with curves corresponding to several different values of C. Using the initial Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
8 Chapter 5 Differential Equations condition (0) = 50, we find the value of C for our specific problem is C = 5. It follows that the solution of the initial value problem is Note (t) = 300 + 5 e -0. t. We could also use the initial condition in equation (3) to solve for C. C 300 300 (t) = - 8 e -0. t C = -8 00 00 300 (t) = + 5 e -0. t 0 40 60 80 t Figure D.9 Figure D.9 also shows this particular solution (in red) among the curves in the general solution. A significant feature of this model is that, for 0 < (0) < 300, the population increases, but not without bound. Instead, it approaches an equilibrium, or steady-state, solution with a value of 300 lim (t) = lim = 300, t t + 5 e-0. t which is the maximum population that the environment (space and food supply) can sustain. This equilibrium population is called the carrying capacity. Notice that all the curves in the general solution approach the carrying capacity as t increases. Quick Quiz. Which of the following is a separable differential equation? (a) y + y' = t (b) y y' = y + t (c) y y' = t csc y Related Exercises 33 34 Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 9. A differential equation of the form g(t) y' (t) = h(t) is (a) first order and separable. (b) second order and separable. (c) non-separable. 3. Which of the following familes of functions satisfies the equation y' = t y? (a) y = e t + C (b) y = t y + C (c) y = C e t 4. The solution of the initial value problem y' = t e -y, y(0) = is (a) y = e t. (b) y = ln t + e. (c) y = -ln e - t. 5. The general solution of the equation k y' + b = 0, for k 0, is a family of curves, all of which are (a) parabolas. (b) exponential curves. (c) lines. 6. The solution of the initial value problem y' - y -3 = 0, y() = is 4 (a) y = 4 t + (b) y = 4 (c) y = 4 t. t + 3 3 4 + 7. The general solution of y' = t y is (a) t + C. 4 (b) t + C. (c) C t... Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
0 Chapter 5 Differential Equations 8. An implicit solution of y' = sec t 4 y 3 + 3 y, y π 4 = is (a) y 4 + y 3 = tan t +. (b) y 4 + y 3 - tan t = 0. (c) y = 4 tan t - y 3. SECTION D.3 EXERCISES Review Questions. What is a separable first-order differential equation?. Is the equation t y' t (t) = + 4 separable? y 3. Is the equation y' (t) = y - t separable? 4. Explain how to solve a separable differential equation of the form g(t) y' (t) = h(t). Basic Skills 5 6. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable. 5. t -3 y' (t) = 6. e 4 t y' (t) = 5 7. 8. d y 3 = t d t y d y = y x + d x 9. y' (t) = e y/ sin t 0. x d w = w (3 x + ), x > 0 d x. x y' (x) = y, x > 0. t + 3 y y' (t) = t y + 4 -y 3 3. y' (t) csc t = 4. y' (t) e t/ = y + 4 5. u' (x) = e x-u Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 6. x u' (x) = u - 4, x > 0 7 6. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. 7. t y' (t) =, y() =, t > 0 8. sec t y' (t) =, y(0) = 9. y y' (t) = 3 t, y(0) = 9 0. y' (t) = e t y, y(0) =. d y = t y +, y() = d t. y' (t) = y 4 t 3 +, y(0) = 4 e t 3. y' (t) =, y(ln ) = y 4. sec x y' (x) = y 3, y(0) = 3 d y 5. = e x-y, y(0) = ln 3 d x 6. y' (t) = cos π y, y() = 4 T 7-3. Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. 7. y' (t) = t y, y() = + x 8. y' (x) = - y, y() = 9. u' (x) = csc u cos x π, u(π) = 30. y y' (x) = x, y() = - + y 3. y' (x) = x + y + 4, y(3) = 5 z + 4 3. z' (x) =, z(4) = x + 6 Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Chapter 5 Differential Equations T T 33. Logistic equation for a population A community of hares on an island has a population of 50 when observations begin (at t = 0). The population is modeled by the initial value problem d = 0.08 -, (0) = 50. dt 00 a. Find and graph the solution of the initial value problem, for t 0. b. What is the steady-state population? 34. Logistic equation for an epidemic When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation d = k -, (0) = 0, d t A where k is a positive infection rate, A is the number of people in the community, and 0 is the number of infected people at t = 0. The model also assumes no recovery. a. Find the solution of the initial value problem, for t 0, in terms of k, A, and 0. b. Graph the solution in the case that k = 0.05, A = 300, and 0 =. c. For a fixed value of k and A, describe the long-term behavior of the solutions, for any 0 with 0 < 0 < A. Further Explorations 35. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The equation u' (x) = x u 7 - is separable. b. The general solution of the separable equation y' (t) = of t. t can be expressed explicitly with y in terms y 7 + 0 y4 c. The general solution of the equation y y' (x) = x e -y can be found using integration by parts. 36 39. Solutions of separable equations Solve the following initial value problems. When possible, give the solution as an explicit function of t. 36. e y ln t y' (t) =, y() = ln t 3 y (y + ) 37. y' (t) =, y() = t cos t 38. y' (t) =, y(0) = - y y + 3 39. y' (t) = 5 t + 6, y() = 0 40 4. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 3 c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. 40. y y' (t) = t + t; y(-) =, y() = 0, y(-) = - 3 4. e -y/ y' (x) = 4 x sin x - x; y(0) = 0, y(0) = ln 4, y π = 0 4. Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses x + y = a. a. Apply implicit differentiation to x + y = a to show that d y - x =. d x y b. The family of trajectories orthogonal to x + y = a satisfies the differential equation d y d x y =. Why? x Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
4 Chapter 5 Differential Equations c. Solve the differential equation in part (b) to verify that y = e C x and then explain why it follows that y = k x. Therefore, the family of parabolas y = k x forms the orthogonal trajectories of the family of ellipses x + y = a. 43. Orthogonal trajectories Use the method in Exercise 4 to find the orthogonal trajectories for the family of circles x + y = a. Applications T T 44. Logistic equation for spread of rumors Sociologists model the spread of rumors using logistic equations. The key assumption is that at any given time, a fraction y of the population, where 0 y, knows the rumor, while the remaining fraction - y does not. Furthermore, the rumor spreads by interactions between those who know the rumor and those who do not. The number of such interactions is proportional to y ( - y). Therefore, the equation that describes the spread of the rumor is y' (t) = k y ( - y), where k is a positive real number. The number of people who initially know the rumor is y(0) = y 0, where 0 y 0. a. Solve this initial value problem and give the solution in terms of k and y 0. b. Assume k = 0.3 weeks - and graph the solution for y 0 = 0. and y 0 = 0.7. c. Describe and interpret the long-term behavior of the rumor function, for any 0 y 0. 45. Free fall An object in free fall may be modeled by assuming that the only forces at work are the gravitational force and air resistance. By Newton's Second Law of Motion (mass acceleration = the sum of external forces), the velocity of the object satisfies the differential equation m v' (t) = m g + f (v), mass acceleration external forces where f is a function that models the air resistance (assuming the positive direction is downward). One common assumption (often used for motion in air) is that f (v) = -k v, where k > 0 is a drag coefficient. a. Show that the equation can be written in the form v' (t) = g - a v k, where a =. m b. For what (positive) value of v is v' (t) = 0? (This equilibrium solution is called the terminal velocity.) c. Find the solution of this separable equation assuming v(0) = 0 and 0 < v g <. a d. Graph the solution found in part (c) with g = 9.8 m s, m =, and k = 0., and verify that the terminal velocity agrees with the value found in part (b). T 46. Free fall Using the background given in Exercise 45, assume the resistance is given by f (v) = -R v, where R > 0 is a drag coefficient (an assumption often made for a heavy medium such as water or oil). a. Show that the equation can be written in the form v' (t) = g - b v, where b R =. m T b. For what value of v 9s v' (t) = 0? (This equilibrium solution is called the terminal velocity.) g c. Find the solution of this separable equation assuming v(0) = 0 and 0 < v <. b d. Graph the solution found in part (c) with g = 9.8 m s, m =, and R = 0., and verify that the terminal velocity agrees with the value found in part (b). 47. Torricelli's law An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli's Law (see figure). If h(t) is the depth of water in the tank, for t 0, then Torricelli's Law implies h' (t) = - k h, where k is a constant that includes g = 9.8 m s, the radius of the tank, and the radius of the drain. Assume that the initial depth of the water is h(0) = H. Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
Section 5.3 Separable Differential Equations 5 a. Find the solution of the initial value problem. b. Find the solution in the case that k = 0. and H = 0.5 m. c. In part (b), how long does it take for the tank to drain? d. Graph the solution in part (b) and check that it is consistent with part (c). T T 48. Chemical rate equations Let y(t) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation d y = -k y n, where k > 0 is a rate constant and the positive integer n is the order of the reaction. d t a. Show that for a first-order reaction (n = ), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction (n = ) assuming y(0) = y 0. c. Graph the concentration for a first-order and second-order reaction with k = 0. and y 0 =. 49. Tumor growth The Gompertz growth equation is often used to model the growth of tumors. Let M (t) be the mass of a tumor at time t 0. The relevant initial value problem is d M = -r M ln M, M (0) = M 0, d t K where r and K are positive constants and 0 < M 0 < K. a. Graph the growth rate function R(M ) = -r M ln M K (which equals M ' (t)) assuming r = and K = 4. For what values of M is the growth rate positive? For what value of M is the growth rate a maximum? b. Solve the initial value problem and graph the solution for r =, K = 4, and M 0 =. Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? c. In the general solution, what is the meaning of K? Additional Exercises T 50. Technology for an initial value problem Solve y' (t) = y e t cos 3 4 t, y(0) =, and plot the solution for 0 t π. 5. Blowup in finite time Consider the initial value problem y' (t) = y n+, y(0) = y 0, where n is a positive integer. a. Solve the initial value problem with n = and y 0 =. b. Solve the initial value problem with n = and y 0 =. c. Solve the problem for positive integers n and y 0 = n -/n. How do solutions behave as t -? Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.
6 Chapter 5 Differential Equations y (y + ) T 5. Analysis of a separable equation Consider the differential equation y' (t) = and carry out the t (t + ) following analysis. a. Show that the general solution of the equation can be written in the form y(t) = C t + - t b. Now consider the initial condition y() = A, where A is a real number. Show that the solution of the initial value problem is y(t) = +A 3 A t t t + - c. Find and graph the solution that satisfies the initial condition y() =. d. Describe the behavior of the solution in part (c) as t increases. e. Find and graph the solution that satisfies the initial condition y() =. f. Describe the behavior of the solution in part (e) as t increases. g. In the cases in which the solution is bounded for t > 0, what is the value of lim t y(t)? T 53. Analysis of a separable equation Consider the differential equation y y' (t) = e t + t and carry out the following analysis. a. Find the general solution of the equation and express it explicitly as a function of t in two cases: y > 0 and y < 0. b. Find the solutions that satisfy the initial conditions y(-) = and y(-) =. c. Graph the solutions in part (b) and describe their behavior as t increases. d. Find the solutions that satisfy the initial conditions y(-) = - and y(-) = -. e. Graph the solutions in part (d) and describe their behavior as t increases.. t. Copyright rinted by Wolfram 04 Mathematica earson Student Education, Edition Inc.