Ordinary Differential Equations: Worked Examples with Solutions Edray Herber Goins Talitha Michal Washington July 31, 2016
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Contents I First Order Differential Equations 5 1 What is a Differential Equation? 7 1.1 Limits and Continuity................................... 7 1.2 Differential Calculus.................................... 9 1.3 Integral Calculus...................................... 11 1.4 Slope Fields......................................... 14 1.5 Classical Mechanics..................................... 15 2 Separable Equations and Population Dynamics 23 2.1 Separable Equations.................................... 23 2.2 Autonomous Equations.................................. 26 2.3 Population Growth Models................................ 29 2.4 Newton s Law of Cooling................................. 37 2.5 Radioactive Decay..................................... 39 3 Exact Equations and Integrating Factors 41 3.1 Exact Equations...................................... 41 3.2 Equidimensional Equations................................ 43 3.3 Bernoulli Equations.................................... 45 3.4 Integrating Factors..................................... 46 3.5 Linear Equations...................................... 47 3.6 Loan Calculations..................................... 48 3.7 Diffusion Problems..................................... 50 3
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Part I First Order Differential Equations 5
Chapter 1 What is a Differential Equation? 1.1 Limits and Continuity Problem 1. Let d 1 be a rational number, and define the sequence of rational numbers x n by x 1 = 1 and x n+1 = 1 + d 1 1 for n = 1, 2,.... Assuming that this sequence converges, show 4 x n the following: a.) 0 x n 1 d 1 for n = 1, 2,... 4 b.) lim x n = 1 + d n 2 Solution. First note that x 1 1 so all terms in the sequence x n 1 0. This gives 1 1 d 1 = = 1 + d 1 ) 1 1 + d 1 ) 1 1 = x n+1 1. x n 4 4 4 As for the value of such a limit, we have the identity ) lim x n+1 = 1 + d 1 n 4 so that if we denote L = lim n x n we have L = 1 + d 1 4 1 L 1 ); lim x n n = L 2 L d 1 4 1 ± d By the quadratic formula, we have two possibilities:. Since d 1, we cannot have the 2 sign or else L would be negative. Hence L = 1 + d. 2 x n = 0. Problem 2. Let θ be an angle between 0 and π radians, and define the five points A = 0, 0), 2 B = 1, 0), C = cos θ, sin θ), D = cos θ, 0), E = cos 2 θ, cos θ sin θ). Note that B and C lie on a circle of radius 1, while D and E lie on a circle of radius cos θ. 7
Let aθ) be the region bounded by the line AB, the line AC, and the arc BC. Let bθ) be the subregion of aθ) which is the triangle ADC. Let cθ) be the subregion of bθ) which is bounded by the line AD, the line AE, and the arc DE. Show the following: a.) b.) area of cθ) area of aθ) = cos2 θ area of bθ) cos θ sin θ = area of aθ) θ c.) cos θ < sin θ θ < 1 cos θ Solution. The area of the sector s of a circle of radius r and an angle θ is s = 1 2 θ r2, so we have area of aθ) = θ 2, cos θ sin θ area of bθ) =, area of cθ) = θ cos2 θ. 2 2 Since cθ) is a subregion of bθ) is a subregion of aθ), we have θ cos 2 θ 2 < cos θ sin θ 2 < θ 2 = cos θ < sin θ θ < 1 cos θ. Problem 3. Using the previous exercise, show that we have the following limits. a.) lim x 0 cos x = 1 b.) lim x 0 sin x x = 1 c.) lim x 0 1 cos x x 2 = 1 2 Solution. First, note that cos x is a continuous function so lim x 0 cos x = cos 0 = 1. Now using the previous exercise with x in place of θ) we have sin x 1 = lim cos x lim x 0 x 0 x lim 1 sin x = 1 = lim x 0 cos x x 0 x = 1. Finally, to work out the third limit, we note that so that we have the limit 1 cos x x 2 = 1 cos2 x x 2 1 + cos x) = 1 1 + cos x 1 cos x 1 lim x 0 x 2 = lim x 0 1 + cos x lim x 0 ) sin x 2 ; x ) sin x 2 = 1 x 2 12 = 1 2. 8
Problem 4. Let F x) be a function satisfying the recursive formula F x + 1) = x F x) where F 1) is nonzero. Show that F x) has a singularity at the nonpositive integers 0, 1, 2,.... Solution. Let k be a nonnegative integer; then k is a nonpositive integer. Using the recursive formula, F x k) = F x k 1) ) F x) = x k) x 1) x k) = F x + 1) x x 1) x k) so that we have the limit F x + 1) lim F x) = lim F x k) = lim x k x 0 x 0 xx 1) x k) = 1)k 1 F 1) lim k! x 0 x which does not converge. Problem 5. Let n be a positive real number. Show that a.) lim x 0 x n log x = 0 b.) lim x x n log x = 0 Solution. We show the first limit. Substitute x = e t/n so that as t = n log x. Hence, as x 0, the variable t. This gives lim x 0 xn log x = 1 n lim t t e t = 0 The value is zero because the exponential function ft) = e t dies faster than the linear function gt) = t grows. As for the second limit, the substitution x = 1/t means t 0 as x so that lim x x n log x = lim t n log t = 0. t 0 1.2 Differential Calculus Problem 6. Let y = fx) be a function which is smooth i.e. a function where derivatives of all order exist. Given a nonegative integer n, define f n) = dn y. Show that for smooth functions fx) n and gx), f g) 0) = f 0) g 0) f g) 1) = f 1) g 0) + f 0) g 1) f g) 2) = f 2) g 0) + 2 f 1) g 1) + f 0) g 2) In general, we can write What are the coefficients C n k,k? f g) n) = n C n k,k f n k) g k). k=0 9
Solution. This exercise shows Leibnitz s Rule: the coefficients ) n n! C n k,k = = k n k)! k! are the Binomial Coefficients. rule. One verifies the formulas above by repeated use of the product Problem 7. Define the function y = cos log x). Find positive constants a, b, and c such that a x 2 y + b x y + c y = 0. Solution. The derivative of y = cos log x is y = sin log x 1 x, so that x y = sin log x. The second derivative is y = cos log x 1 x ) 1 sin log x 1x ) x 2 and so x 2 y = x y y. Hence, we may choose a = b = c = 1. = x 2 y = sin log x cos log x Problem 8. For real numbers x 1, x 2,..., x n and positive integers C 1, C 2,..., C n ; define the polynomial gx) = x x 1 ) C 1 x x 2 ) C2 x x n ) Cn. Show the identity Solution. The logarithm of gx) is the sum which in turn has the derivative g x) gx) = C 1 + C 2 + + C n. x x 1 x x 2 x x n log gx) = C 1 logx x 1 ) + C 2 logx x 2 ) + + C n logx x n ) g x) gx) = d [ log gx) ] = C 1 1 x x 1 + C 2 1 x x 2 + + C n 1 x x n. Problem 9. Denote the curve C = { x, y) R 2 y 3 = x 5 1 }. Without solving for y, show that show that a.) 1, 0) C has a vertical tangent. b.) 0, 1) C has a horizontal tangent. 10
c.) The tangent line at every other point on C has positive slope. Solution. The curve C is defined by y 3 x 5 + 1 = 0, so we can find the derivative via implicit differentiation: 3 y 2 dy 5 dy x4 = 0 = = 5 x4 3 y 2. When x = 1 and y = 0, the slope dy/ = so the curve has a vertical tangent. In fact, when y = 0, then x 5 = 1 and the only real number where this is true is x = 1. That is, 1, 0) is the only point on C with a vertical tangent. When x = 0 and y = 1, the slope dy/ = 0 so the curve has a horizontal tangent. In fact, when x = 0, then y 3 = 1 and the only real number where this is true is y = 1. That is, 0, 1) is the only point on C with a horizontal tangent. Otherwise both x 4 and y 2 are nonzero and hence must be positive, so that dy/ is positive as well. Hence the slope is always positive in these cases. Problem 10. Let y = yx) be any function which satisfies the implicit relation x = tan y. a.) Show that y = yx) must be an increasing function without relative extrema. b.) It is true that y0) = 0 and y 0) = 1? Solution. First we make an observation: 1 + x 2 = 1 + tan 2 y = cos2 y + sin 2 y cos 2 y = 1 cos 2 y = sec2 y. Now we implicitly differentiate the expression tan y x = 0: sec 2 y dy 1 = 0 = dy = cos2 y = 1 1 + x 2. Hence dy/ 0 so that it has no relative extrema. In fact, dy/ > 0 so that it is always increasing. Note that y 0) = 1/1 + 0 2 ) = 1 must indeed be true. It need not be true that y0) = 0, but it must be true that y0) satisfy tan y0) = 0. This occurs only when y0) = n π where n is an integer. 1.3 Integral Calculus Problem 11. Compute the value of the following integrals: a.) b.) c.) 1 1 2 0 5 3 1 x 2 2 x x 2 x 3)5 x) 11
Solution. If x = a + b sin θ then = b cos θ dθ. This yields 1 1 π/2 = 1 x 2 π/2 cos θ dθ π/2 1 sin 2 θ = π/2 dθ = π. With the substitution x = 1 + sin θ, as x ranges from 0 to 2, the function sin θ ranges from -1 to +1, so the variable θ ranges from π/2 to π/2. Through the identity 2 x x 2 = 1 x 2 2 x + 1) = 1 x 1) 2 = 1 sin 2 θ = cos 2 θ we may evaluate the integral as 2 0 π/2 = 2 x x 2 π/2 cos θ dθ π/2 cos 2 θ = π/2 dθ = π. Again, with the substitution x = 4 + sin θ, as x ranges from 3 to 5, the function sin θ ranges from -1 to +1, so the variable θ ranges from π/2 to π/2. Through the identity x 3)5 x) = 1 + sin θ)1 sin θ) = 1 sin 2 θ = cos 2 θ we may evaluate the integral as 5 3 x 3)5 x) = π/2 π/2 dθ = π. Problem 12. Given that a x 2 +b x+c is a quadratic polynomial with leading term a < 0, discriminant b 2 4 a c > 0, and roots α < β, show that β α a x 2 + b x + c = π a 1/2. Solution. We ll make the substitution x = b b 2 a + 2 4 a c b sin θ = = 2 4 a c cos θ dθ. 2 a 2 a First, using the quadratic formula to find the roots of a x 2 + b x + c, we define α = b b 2 a 2 4 a c, β = b b 2 a 2 a + 2 4 a c 2 a so that as x ranges from α to β, the function sin θ ranges from -1 to +1, and hence the variable θ ranges from π 2 to +π. We also have the identity remember that a < 0): 2 a x 2 + b x + c = a so that the integral is [ b 2 4 a c 4 a 2 = [ b 2 ] 1/2 4 a c a 4 a 2 cos 2 1/2 θ = a dθ β α x + b ) ] 2 1/2 = [ b 2 ] 1/2 4 a c a 2 a 4 a 2 b2 4 a c 4 a 2 sin 2 θ a x 2 + b x + c = 1 π/2 a 1/2 dθ = π/2 π a 1/2. 12
Problem 13. Consider the identity Compute the value of this sum/integral. 1 1 3 + 1 5 1 7 + 1 9 1 11 + = 1 0 x 2 + 1. Solution. We ll substitute x = tan θ. As x ranges from 0 to 1, the variable θ ranges from 0 to π/4. Recall that the derivative of tan θ is sec 2 θ; hence = sec 2 θ dθ. Also, we have the identity so that the integral becomes x 2 + 1 = tan 2 θ + 1 = sin2 θ cos 2 θ + 1 = sin2 θ + cos 2 θ cos 2 θ 1 0 π/4 x 2 + 1 = sec 2 θ dθ 0 sec 2 θ = π/4 0 = 1 cos 2 θ = sec2 θ dθ = π 4. Problem 14. Show that when a > b > 0, 1 + a 2 x 2 ) 1 + b 2 x 2 ) = π a + b. Solution. First consider the partial fraction expansion 1 1 + a 2 x 2 ) 1 + b 2 x 2 ) = A 1 + a 2 x 2 + B 1 + b 2 x 2 where A and B satisfy the equations b 2 A + a 2 B = 0 and A + B = 1. That is, A = a 2 /a 2 b 2 ) and B = b 2 /b 2 a 2 ). The integral of interest can now be expressed as the sum of two integrals: 1 + a 2 x 2 ) 1 + b 2 x 2 ) = A 1 + a 2 x 2 + B 1 + b 2 x 2. Now let s focus on the integral /1 + a2 x 2 ). We ll make the substitution x = 1 tan θ. As x a ranges from to, the variable θ ranges from π/2 to π/2. This gives π/2 1 + a 2 x 2 = 1 π/2 a Finally, we can evaluate the original integral. It is 1 + a 2 x 2 ) 1 + b 2 x 2 ) = A π a + B π b sec 2 θ π/2 1 + tan 2 θ dθ = 1 π/2 a dθ = π a. = a π a 2 b 2 + b π a b) π b 2 = a2 a 2 b 2 = π a + b. 13
Problem 15. Given that a x 2 + b x + c is a positive definite quadratic polynomial that is, a > 0 and b 2 4 a c < 0), show that Solution. We ll make the substitution a x 2 + b x + c = 2π 4 a c b 2. x = b 2 a + 4 a c b 2 2 a tan θ = = 4 a c b 2 2 a sec 2 θ dθ. As x ranges from to, the variable θ ranges from π/2 to +π/2. Using the identity tan 2 θ+1 = sec 2 θ, we also have the identity [ a x 2 + b x + c = a x + b ) ] 2 + 4 a c b2 2 a 4 a 2 [ 4 a c b 2 = a 4 a 2 tan 2 θ + 4 a c ] b2 4 a 2 = 4 a c b2 4 a so that the integral is π/2 a x 2 + b x + c = π/2 2 4 a c b 2 dθ = sec 2 θ = 2π 4 a c b 2. 4 a c b 2 2 dθ 1.4 Slope Fields For the following differential equations, draw a slope field for the given differential equation and state whether you think that the solutions are converging or diverging. Problem 16. dy = x y + 0.1 y3. Solution. Figure 1.1 contains a plot of the slope field. Convergence of the solution depends on the initial condition. The graph shows that if y0) 2.4 then y ±, yet if y0) 2.2 then y 0. Figure 1.2 contains a closer analysis of the direction field. Using this, we see that the solutions diverge y ± if y0) > 2.37, yet the solutions converge y 0 if y0) < 2.37. Problem 17. dy = x2 + y 2. Solution. Figure 1.3 contains a plot of the slope field. It is clear that the solutions are diverging y regardless of the initial condition. 14
Figure 1.1: Direction Field for dy = x y + 0.1 y3 4.8 4 3.2 2.4 1.6 0.8-0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-0.8 1.5 Classical Mechanics Problem 18. At t = 0, a ball is propelled downward from an initial height of 1000 m with an initial speed of 25 m/s. Calculate the time t that the ball hits the ground. Solution. The height of the ball at time t is given by the formula xt) = x 0 + v 0 t 1 2 g t2 where x 0 = 1000 m is the initial height, v 0 = 25 m/s is the initial speed note negative because the ball is propelled downward!), and g = 9.81 m/s 2 is the acceleration due to gravity. We want to find the time t when xt) = 0 i.e. 0 = 1000 25 t 1 2 9.8 t2 = t = 11.9556 or 17.0525. Time can only be positive, so we find t = 12 s. Problem 19. A balloon is ascending at a rate of 15 m/s at a height of 100 m about the ground, when a package is dropped from the gondola. How long will it take the package to reach the ground? Ignore air resistance. 15
Figure 1.2: Solutions of dy = x y + 0.1 y3, with y0) = 2.37 3.5 3.25 3 2.75 2.5 2.25 0 0.08 0.16 0.24 0.32 0.4 0.48 0.56 0.64 0.72 0.8 0.88 0.96 1.04 Solution. Denote the height of the package at time t by xt) = x 0 + v 0 t 1 2 g t2 where x 0 = 100 m is the initial height, v 0 = 15 m/s is the initial velocity note positive because the balloon is ascending!) and g = 9.81 m/s 2 is the acceleration due to gravity. We want to find the time t when xt) = 0 i.e. 0 = 100 + 15 t 1 2 9.81 t2 = t = 6.29616 or 3.23806. Time can only be positive, so we find t = 6.3 s. Problem 20. The velocity vt) of a falling object satisfies the initial value problem dv = 9.8 v 5 where v0) = 0. a.) Find the time that must elapse for the object to reach 98% of its terminal velocity. b.) How far does the object fall in the time found in part a)? Solution. The velocity of the falling object at t seconds is vt) = 49 1 e t/5). The limiting 16
Figure 1.3: Direction Field for dy = x2 + y 2 0.75 0.5 0.25-1.25-1 -0.75-0.5-0.25 0 0.25 0.5 0.75 1 1.25-0.25-0.5-0.75 velocity is v L = 49 m/sec, so the velocity reaches 98% of this value when 49 1 e t/5) = 0.98 49 1 e t/5 = 0.98 e t/5 = 0.02 e t/5 = 50 = t = 5 ln 50 = 19.56 sec. The distance the object has fallen after t seconds is xt) = 49 t+245 e t/5 245. Hence the distance after t = 19.56 seconds is x = 49 19.56 + 245 0.02 245 = 718.346 m. Problem 21. A skydiver of mass 60 kg free-falls from an airplane at an altitude of 5000 meters. He is subjected to an air resistance force that is proportional to his speed. Assume that the constant of proportionality is 10 kg/sec). Find and solve the differential equation governing the altitude of the skydiver at time t seconds after the start of his free-fall. Assuming that he does not employ his parachute, find his limiting velocity and how much time will elapse before he hits the ground. Solution. Let xt) denote the altitude of the skydiver at time t, and vt) = the velocity of the skydiver at time t. Then when the skydiver initially jumps from the plane i.e. at t 0 = 0 we have xt 0 ) = 5000 m and vt 0 ) = 0 m/s. The force due to gravity is m g = 588.6 kg m/s 2, and the force due to air resistance is 10 v. Hence the equation of motion is m d2 x = m g 10 2. 17
First we solve for the velocity. We have the initial value problem This equation is separable: 60 dv = 588.6 10 v where v0) = 0. dv v + 58.86 = 6, so that ln v + 58.86 = t 6 + C 1. The constant is C 1 = ln 58.86, so we have ln v + 58.86 = t 6 + ln 58.86, which gives vt) = 58.86 e t/6 58.86. As t, the limiting velocity is vt) 58.86 m/s. Second we solve for the altitude. We have the initial value problem = 58.86 e t/6 58.86 where x0) = 5000. We integrate term by term to find xt) = 58.86 6 e t/6 58.86 t+c 2. Using the initial conditions, we find that the constant is C 2 = 5353.16, so we have xt) = 5353.16 58.86 t + 353.16 e t/6. To find out how much time will elapse before the skydiver hits the ground, we want to find the time t such that t = 0, i.e., 0 = 5353.16 58.86 t + 353.16 e t/6 so that t = 90.9473 s. Problem 22. A sky diver weighing 180 lb including equipment) falls vertically downward from an altitude of 5,000 ft and opens the parachute after 10 sec of free fall. Assume that the force of air resistance is 0.75 v when the parachute is closed and 12 v when the parachute is open, where the velocity v is measured in ft/sec. Note: The force due to gravity is m g = 180 lb where the gravitational acceleration is g = 32 ft/sec 2. The mass of the sky diver is m = 180/g = 5.63 lb. a.) Find the speed of the sky diver when the parachute opens. b.) Find the distance fallen before the parachute opens. c.) What is the limiting velocity v L after the parachute opens? d.) Determine how long the sky diver is in the air after the parachute opens. e.) Plot the graph of velocity versus time from the beginning of the fall until the skydiver reaches the ground. Solution. Let v = vt) denote the speed of the sky diver in the downward direction. The force due to air resistance is γ v, where γ = { 0.75 lb/sec when the parachute is closed, 12 lb/sec when the parachute is open. Now we find an expression for the speed of the sky diver. Newton s Second Law of Motion states F = m a, which is equivalent to the differential equation m dv = m g γ v. We solve this equation 18
using integrating factors. After dividing through by m, we multiply by µt) = e γt/m : γt/m dv e m dv + γ v = m g dv + γ m v = g + γ m eγt/m v = g e γt/m d [ ] e γt/m v = g e γt/m = e γt/m vt) = m g γ eγt/m + C for some constant C. Say that we have the initial condition vt 0 ) = v 0. If we set t = t 0, we have vt 0 ) = m g γ + C e γt 0/m = C = vt 0 ) m g ) e γt0/m. γ Hence the speed of the sky diver is vt) = m g γ + C e γt/m = m g γ + v 0 m g ) e γt t0)/m. γ Initially the sky diver is in free fall, so v 0 = 0 ft/sec when t 0 = 0 sec. The speed of the sky diver when the parachute opens is v10) = 180 0.75 + 0 180 ) e 0.7510 0)/5.63 = 176.74 ft/sec. 0.75 Let x = xt) denote the distance the sky diver has fallen by time t. Since we have chosen v = vt) as the velocity in the downward direction, we have the initial value problem = v where xt 0 ) = x 0. We have the differential equation = m g γ + v 0 m g ) γt t e 0)/m = xt) = m g γ γ t v 0 m g ) m γ γ e γt t 0)/m + C for some constant C. Upon setting t = t 0 we have x 0 = m g γ t 0 v 0 m g γ Hence the distance the sky diver has fallen is ) m γ + C = C = x 0 m g γ t 0 + xt) = x 0 + m g γ t t 0) + v 0 m g γ ) m γ 1 e γt t 0)/m ). v 0 m g ) m γ γ. Initially we have v 0 = 0 ft/sec and x 0 = 0 ft when t 0 = 0 sec, so the distance the sky diver has fallen when the parachute opens is x10) = 0 + 180 10 0) + 0 180 ) 5.63 1 e 0.7510 0)/5.63) = 1074.47 ft. 0.75 0.75 0.75 19
After the parachute opens, the differential equation which governs the motion of the sky diver remains the same except that γ = 12 lb/sec. Hence the speed of the sky diver is vt) = m g γ + v 0 m g ) e γt t 0)/m γ where now v 0 = 176.74 ft/sec when t 0 = 10 sec. Hence the limiting velocity after the parachute opens is v L = lim vt) = m g t γ = 180 = 15 ft/sec. 12 After the parachute opens, the distance the sky diver has fallen is xt) = x 0 + m g γ t t 0) + v 0 m g ) ) m 1 e γt t 0)/m γ γ where now v 0 = 176.74 ft/sec and x 0 = 1074.47 ft when t 0 = 10 sec. Say that T = t t 0 is the time that the sky diver is in the air after the parachute opens i.e., xt) = 5000 ft. If T is large enough, then e γt/m 0, so we can ignore it. Hence we have the equation so that T = 256.65 sec. xt + t 0 ) x 0 + m g γ T + 5000 1074.47 + 180 12 T + 3849.71 15 T v 0 m g γ ) m γ 176.74 180 12 ) 5.63 12 Figure 1.4: Plot of vt) vs. t 175 150 125 100 75 50 25-50 -25 0 25 50 75 100 125 150 175 200 225 250-25 20
Figure 1.4 has a plot of the function: 180 0.75 + vt) = 0 180 0.75 ) e 0.75t/5.63 180 12 + 176.74 180 ) e 12t 10)/5.63 12 if t 10 sec, if t 10 sec. Problem 23. For small, slowly falling objects, the assumption that the drag force is proportional to the velocity is a good one. For larger, more rapidly falling objects, it is more accurate to assume that the drag force is proportional to the square of the velocity. a.) Write a differential equation for the velocity of a falling object of mass m if the drag force is proportional to the square of the velocity. b.) Determine the limiting velocity after a long time. c.) If m = 10 kg, find the drag coefficient so that the limiting velocity is 49 m/sec. d.) Using the data in part c), draw a slope field. Solution. Let v = vt) denote the velocity of the falling object in the downward direction at time t. The force due to gravity on this mass is m g. The force due to air resistance is proportional to v 2, so this force is k v 2 for some positive constant k. Newton s Second Law of Motion states that F = m a is the sum of the forces on this object, so we have the differential equation m dv = m g k v2. Let v L = lim vt) denote the limiting velocity after a long time. It satisfies the equation m g k v 2 L = 0, t so that v L = m g k. Hence v m g k. Gravitational acceleration is g = 9.8 m/sec2. If the limiting velocity is to be v L = 49 m/sec for a mass m = 10 kg, then we have m g k v 2 L = 0, so that k = m g 10 9.8 kg m/sec 2 = v 2 L 49 2 m 2 /sec 2 = 2 49 kg/m. Hence k = 2/49 = 0.041 kg/m. We consider the differential equation m dv = m g k v2, which we can express as dv = 9.8 0.0041 v2. A plot of the direction field can be found in Figure 1.5, along with the equilibrium solution v = 49 m/sec. 21
Figure 1.5: Slope Field for dv = 9.8 0.0041 v2 50 25 0 25 50 75 100 125-25 22
Chapter 2 Separable Equations and Population Dynamics 2.1 Separable Equations Problem 1. Find the general solution of the following differential equation. If possible, find an explicit solution: dy = 1 + y 2) e x. Solution. This is a separable equation: dy = 1 + y 2) e x 1 1 + y 2 dy = ex 1 1 + y 2 dy = e x = arctan y = e x + C. Hence the solution to the differential equation is yx) = tan e x + C). Problem 2. Find the exact solution of the following initial value problem. Indicate the interval of existence. dy = 1 + y 2) where y0) = 1. Solution. This equation is separable: dy = 1 + y2 dy 1 + y 2 = arctan y = x + C = yx) = tan x + C). 23
When x = 0 we want y = 1, so we choose C such that tan C = 1. This happens when C = π 4 modulo multiples of π) so yx) = tan x + π ). Now tan C is defined on the interval π 4 2, π ) 2 notice that C = π 4 is in this interval so we must have π 2 < x + π 4 < π 2 3 π 4 < x < π. Hence the interval of existence is 3 π 4 4, π ). 4 which implies Problem 3. Solve the initial value problem dy = 2 1+x) 1+y2 ) subject to y0) = 0 and determine where the solution attains its minimum value. Solution. This differential equation is a separable equation: dy = 2 1 + x) 1 + y 2) 1 dy 1 + y 2 = 2 + 2 x d [arctan y] = 2 + 2 x = arctan y = x2 + 2 x + C for some constant C. Hence yx) = tan x 2 + 2 x + C ) is the general solution. When x = 0, we find that 0 = y0) = tan C, so that C must be an integer multiple of π. Using the Angle Addition Formula for tangent: tan θ + tan C tan θ + C) = 1 tan θ tan C = tan θ we have the explicit solution yx) = tan x 2 + 2 x ). This solution attains its minimum value when its derivative vanishes. But y = 2 1 + x) 1 + y 2 ), and 2 1 + y 2 ) > 0, so this happens when 1 + x = 0. Hence the solution attains its minimum value when x = 1. Problem 4. Consider the differential equation dy = 1 2 x. y a.) Find the solution y = yx) corresponding to y1) = 2. b.) Plot the graph of the solution. c.) Determine the interval in which the solution is defined. Solution. This differential equation is a separable equation: dy = 1 2 x y y dy = 1 2 x ) 1 ) y dy = 2 x = y 2 2 = x x2 + C 24
for some constant C. Upon evaluating at the point x 0, y 0 ) = 1, 2) we find that C = 2. Hence the solution to the initial value problem is yx) = 2 x 2 x 2 + 4. Figure 2.1 has a plot of the solution. The solution y = 2 2 x) 1 + x) is defined when the factors 2 x and 1 + x have the same sign, so the solution is defined when 1 < x < 2. Figure 2.1: Plot of yx) = 2 x 2 x 2 + 4 0.5-2 -1.5-1 -0.5 0 0.5 1 1.5 2 2.5-0.5-1 -1.5-2 -2.5 Problem 5. Solve the given initial value problem dy = 4 x y where y0) = y 0. Determine how the interval in which the solution exists depends on the initial value y 0. Solution. This differential equation is a separable equation: dy = 4 x y y dy = 4 x y dy = 4 x = y 2 2 = 2 x2 + C for some constant C. Upon evaluating at the point t 0, y 0 ) = 0, y 0 ), we find that C = y 0 2 we find the expression 2. Hence y 2 2 = 2 x2 + y 0 2 2 = yx) = ± y 0 2 4 x 2. 25
This solution exists when x < 1 2 y 0. In particular, no solution exists if y 0 = 0, so we must also have y 0 0. 2.2 Autonomous Equations In the three exercises below, an autonomous differential equation is given in the form dy = fy). Perform each of the following tasks without the aid of technology. a.) Sketch a graph of fy). b.) Use the graph of f to develop a phase line for the autonomous equation. equilibrium point as either unstable or asymptotically stable. Classify each c.) Sketch the equilibrium solutions in the xy plane. These equilibrium solutions divide the xy plane into regions. Sketch at least one trajectory in each of these regions. Problem 6. dy = y + 1) y 4). Solution. Figure 2.2 has a graph of fy) = y + 1) y 4). The equilibrium solutions are y = 1 and y = 4. Figure 2.3 has a graph of the phase line. We use the First Derivative Test to check for the stability: since fy) = y + 1) y 4) = y 2 3 y 4 we have df = 2 y 3. When y = 1 we dy have df df = 5 is negative, so that y = 1 is stable. When y = 4 we have = 5 is positive, so dy dy that y = 4 is unstable. Figure 2.4 has a graph of the trajectories in the various regions. Problem 7. dy = 6 + y y2. Solution. We have the factorization fy) = 6 + y y 2 = y + 2) y 3). Figure 2.5 has a graph of fy). The equilibrium solutions are y = 2 and y = 3. Figure 2.6 has a graph of the phase line. We use the First Derivative Test to check for the stability: since fy) = 6 + y y 2 we have df dy we have df = 1 2 y. When y = 2 we have df dy = 5 is positive, so that y = 2 is unstable. When y = 3 = 5 is negative so that y = 3 is stable. Figure 2.7 has a graph of the trajectories in dy the various regions. Problem 8. dy = y y 1) y 2). Solution. Figure 2.8 has a graph of fy) = y y 1) y 2). The equilibrium solutions are y = 0, y = 1, and y = 2. Figure 2.9 has a graph of the phase line. We use the First Derivative Test to check for the stability: since fy) = y 3 3 y 2 +2 y we have f y) = 3 y 2 6 y +2. The values of interest are 26
Figure 2.2: Plot of fy) = y + 1) y 4) f 0) = 2 > 0, f 1) = 1 < 0 and f 2) = 2 > 0. Hence y = 0 and y = 2 are unstable equilibria, whereas y = 1 is a stable equilibrium. Figure 2.10 has a graph of the trajectories in the various regions. Problem 9. Draw a slope field for the differential equation dy = 3 2 y. Use this to determine the behavior of y as x. Does this behavior depends on the initial condition of y at x = 0? Solution. Figure 2.11 contains a plot of the slope field for the differential equation. It is clear that y 3 2 as x, regardless of the initial condition of y. Problem 10. Sometimes a constant equilibrium solution has the property that solutions lying on 27
Figure 2.3: Phase Line for dy = y + 1) y 4) one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable. a.) Consider the equation dy = k 1 y)2, where k is a positive constant. Show that y L = 1 is the only critical point. b.) Sketch fy) versus y. Show that y is increasing as a function of x for y < 1 and also for y > 1. Note: The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore y L = 1 is semistable. c.) Solve dy = k 1 y)2 subject to the initial condition y0) = y 0 and confirm the conclusions reached in part b). Solution. Denote the function fy) = k 1 y) 2. The critical points y L correspond to fy L ) = 0, so y L = 1 is the only critical point. Figure 2.12 contains a plot of fy) = k 1 y) 2 versus y. Say that y = yx) is a solution to the differential equation y = fy). When y 1 we have fy) > 0, so the derivative y x) is positive. Hence y is an increasing function of x whenever y 1. To solve the differential equation, divide both sides by 1 y) 2 : dy = k 1 y)2 y 1) 2 dy = k y 1) 2 dy = k = y 1) 1 = k t + C for some constant C. When t = 0, we have C = y 0 1) 1. We may use this to explicitly solve for y as a function of time: 1 y 1 = k t 1 y 0 1 1 y 1 = y 0 1) k t 1 y 0 1 y 1 = y 0 1 1 + 1 y 0 ) k t = yt) = 1 + y 0 1 1 + 1 y 0 ) k t = y 0 + 1 y 0 ) k t 1 + 1 y 0 ) k t. 28
Figure 2.4: Plot of solutions to dy = y + 1) y 4) If y 0 < 1 then 1 y 0 ) > 0 so that the denominator for y = yt) does not vanish for t 0. As t increases without bound, we see that y 1. On the other hand, if y 0 > 1 then 1 y 0 ) < 0 so that the denominator for y = yt) vanishes when t = 1/ y 0 1) k) > 0. Hence there is a finite value of t for which y. We conclude that solutions y < 1 approach the limiting solution y L = 1, yet solutions y > 1 grow farther away. 2.3 Population Growth Models Problem 11. The size P t) at t months of a certain population satisfies the differential equation dp = 0.5 p 450. a.) Find the time at which the population becomes extinct if P 0) = 850. b.) Find the time of extinction if P 0) = P 0, where 0 < P 0 < 900. 29
Figure 2.5: Plot of fy) = 6 + y y 2 c.) Find the initial population P 0 if the population is to become extinct in 1 year. Solution. The solution so the differential equation is P t) = 900 + C e t/2 for some constant C. If P 0) = 850, then C = 50. The population becomes extinct when 0 = P t) = 900 50 e t/2. This gives e t/2 = 900 = 18, so that the population becomes extinct when t = 2 ln 18 = 5.78074 months. 50 More generally, if P 0) = P 0, then C = P 0 900. The population becomes extinct when 0 = P t) = 900 + P 0 900) e t/2. This gives e t/2 900 =, so that the population becomes extinct when 900 P 0 900 t = 2 ln months. If the population is to become extinct in 1 year, then P 12) = 0. This 900 P 0 gives 900 + P 0 900) e 12/2 = 0, so that P 0 = 900 e 6 1)/e 6 = 897.769. Problem 12. Suppose that you have a closed system containing 1000 individuals. A flu epidemic starts. Let Nt) represent the number of infected individuals in the closed system at time t. Assume 30
Figure 2.6: Phase Line for dy = 6 + y y2 that the rate at which the number of infected individuals is changing jointly proportional to the number of infected individuals and the number of noninfected individuals. Furthermore, suppose that when 100 individuals are infected, the rate at which individuals are becoming infected is 90 individuals per day. If 20 individuals are infected at time t = 0, when will 90% of the population be infected? Solution. Assuming that there are only healthy individuals and sick individuals, if Nt) is the number of sick, then 1000 Nt) is the number of healthy. Then there is a constant of proportionality k such that dn = k Nt) 1000 Nt)). When Nt) = 100, we want N t) = 90. Hence, k = 90/ 100 900) = 1/1000 day 1. The differential equation is dn = 1 1000 N 1000 N) = Nt) = 1 1/1000 + C e t where the time t is measured in days. If N0) = 20, then we have C = 49/1000. Hence Nt) = 1000 1 + 49 e t. We want to find the number of days t such that Nt) = 90% 1000 = 900 individuals: 1000 1 + 49 e t = 900 = et = 441 = t = 6.08904 Hence 90% of the population will be infected after approximately 6 days. Problem 13. A population of bacteria, growing according to the Malthusian model, doubles itself in 10 days. If there are 1,000 bacteria present initially, how long will it take the population to reach 10,000? Solution. Let P t) denote the number of bacteria present at time t days. The Malthusian model states that there are constant P 0 and r such that P t) = P 0 e rt. Initially, there are 1000 bacteria present, so P 0 = 1000. The population doubles itself in 10 days, so when t = 10 we have P t) = 2000. Hence 2000 = 1000 e 10r so that r = ln 2 10 a population of 10,000 satisfies 10000 = 1000 e rt, so t = 33.2 days. 31 = 0.0693147. The time t it will take to reach ln 10 r = 33.2193. Hence it will take
Figure 2.7: Plot of solutions to dy = 6 + y y2 Problem 14. Suppose a population is growing according to the logistic equation dp = r P 1 P ). K Prove that the rate at which the population is increasing is at its greatest when the population is at one-half of its carrying capacity. Solution. The rate at which the population P is increasing is fp ) = r P 1 P ) = r P r K K P 2. The rate at which the rate is increasing is df = df dp dp = [ r 2 r K P ] [r P r K P 2] = r 2 P 32 1 2 P K ) 1 P ). K
Figure 2.8: Plot of fy) = y y 1) y 2) 0.75 0.5 f'y)>0 f'y)<0 f'y)>0 0.25-0.25 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 y=0 y=1 y=2-0.25-0.5 0.75-0.75 0.5 f'y)>0 0.25 f'y)<0 Figure 2.9: Phase Line for dy = y y 1) y 2) f'y)>0-0.25 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 y=0 y=1 y=2-0.25 We find relative extrema -0.5 when P = 0, K 2, and K; which gives the values fp ) = 0, r K 4, and 0, respectively. When P = 0, K the rate fp ) = 0; but when P = K/2 the rate fp ) > 0 is increasing. Hence -0.75 the rate at which the populaiton is increasing at the greatest is when P = K 2, i.e., when the population is one-half its carrying capacity. Problem 15. A population is observed to obey the logistic equation with the eventual population 20,000. The initial population is 1,000, and 8 hours later, the observed population is 1,200. Find the reproductive rate and the time required for the population to reach 75% of its carrying capacity. Solution. Let P t) denote the size of the population at time t hours. This function satisfies the differential equation dp = r P 1 P ) K P 0 20000 = P t) = = K P 0 + K P 0 ) e rt 1 + 19 e rt, 33
Figure 2.10: Plot of solutions to dy = y y 1) y 2) 3 2.5 2 1.5 1 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 where r is the reproductive rate, P 0 = 1, 000 is the initial population size, and K = 20, 000 is the eventual population size. When t = 8 we know that P t) = 1200. We find that 1200 = 20000 1 + 19 e 8r = e 8r = 57 47 The time it will take to reach 75% of its carrying capacity is = r = 0.024113. 75% 20000 = 20000 1 + 19 e rt = e rt = 57 = t = 167.671 Hence the time is 167.7 hours. Problem 16. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population P : The more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by E P, where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by dp = r 1 P K ) P E P. a.) Show that if E < r, then there are two equilibrium points, namely P 1 = 0 and P 2 = K 1 E ) > 0. r 34
Figure 2.11: Slope Field for dy = 3 2 y 2 1.5 1 0.5-0.5 0 0.5 1 1.5 2 2.5 3 3.5 4-0.5-1 b.) Show that P = P 1 is an unstable equilibrium and P = P 2 is a stable equilibrium. c.) A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort E and the asymptotically stable population P 2. Find Y as a function of the effort E; the graph of this function is known as the yield-effort curve. d.) Determine E so as to maximize Y and thereby find the maximum sustainable yield Y m. Solution. Consider the function fp ) = r 1 P K ) y E P = rk P 2 + r E) P = rk P P + ) K E r). r K E r) Hence P = 0 and P = = K 1 E ) are equilibrium points. If E < r then r r 1 E r > 0.) We have the derivative f P ) = 2 r K P + r E), which shows f P 1 ) = r E > 0 yet f P 2 ) = E r < 0. Hence P = P 1 is an unstable equilibrium whereas P = P 2 is a stable equilibrium. The sustainable yield is Y E) = E K 1 E ) = K r r E2 + K E. This function has a maximum when Y E) = 2 K r E + K = 0, so that E = r. Hence the maximum sustainable 2 yield is Y m = K r 1 1 ) = K r 2 2 4. 35
Figure 2.12: Plot of fy) = k 1 y) 2 1 0.75 0.5 0.25 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 y=1-0.25-0.5 Problem 17. Some diseases such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease but who exhibit no overt symptoms. Let S and I, respectively, denote the proportion of susceptibles and carriers i.e., infected) in the population, respectively. Suppose that carriers are identified and removed from the population at a rate β, so di also that the disease spreads at a rate proportional to the product of S and I; thus ds = β I. Suppose = α S I. a.) Determine It) at any time t by solving di = β I subject to the initial condition I0) = I 0. b.) Use the result of part a) to find St) at any time t by solving ds = α S I subject to the initial condition S0) = S 0. c.) Find the proportion of the population that escapes the epidemic by finding the limiting value of St) as t. Solution. To determine It), we solve a separable equation: di = β I 1 di = β I 1 I di = β ln I = β t + C 1 = It) = C e βt 36
where C = ±e C 1 is a constant. When t = 0 we have C = I 0, so that It) = I 0 e βt. To determine St), we solve another separable equation: where C = ±e C 1 ds = α S I 0 e βt 1 S ds = α I 0 e βt 1 S ds = α I 0 e βt ln S = α I 0 β e βt + C 1 = St) = C exp is a constant. When t = 0, we have [ C = S 0 exp α I ] 0 β [ α I0 β e βt 1 e = St) = S 0 exp [ α βt ] I 0. β ] As time increases without bound, we have [ lim St) = S 0 exp α I ] 0. t β 2.4 Newton s Law of Cooling Problem 18. Newton s law of cooling asserts that the rate at which an object cools is proportional to the difference between the object s temperature T ) and the temperature of the surrounding medium T ). Let s suppose that the ambient temperature is T = 70 F and that the rate constant is 0.05 min) 1. Write a differential equation for the temperature of the object at any time. Solution. Let T t) denote the temperature of the object in F at any time t in minutes. Newton s law of cooling states that dt is proportional to T 70 F, so say that dt = r T 70 F) for some constant r. We know that r = 0.05 min) 1, so we determine the sign of r. If T > 70 F then the object will cool i.e., dt < 0. Hence r must be a negative constant. The differential equation must be dt = 0.05 T 70). Problem 19. Assuming Newton s Law of Cooling, show that T t) = T + T 0 T ) e rt, where T 0 is the temperature of the body at time t = 0 and r is the proportionality constant. Solution. Newton s law of cooling is equivalent to the differential equation T T T ), i.e., T = r T T ). Note that if T > T then the object will cool, i.e., T < 0, whereas if T < T 37
then the object will warm up, i.e., T > 0. Hence r is a negative constant of proportionality. This equation is separable: dt = r T T ) dt T T = k ln T T ln T 0 T = r ln T T T 0 T = r t T T T 0 T = ert = T T T 0 T = ±e rt When t = 0 we see that the left-hand side is 1, so we must choose the positive sign on the right-hand side. Hence T t) T T 0 T = e rt, so that T t) = T + T 0 T ) e rt. Problem 20. A murder victim is discovered at midnight and the temperature of the body is recorded at 31 C. One hour later, the temperature of the body is 29 C. Assume that the surrounding air temperature remains constant at 21 C. Calculate the victim s time of death. Note: The normal temperature of a living human being is approximately 37 C. Solution. Assume that the temperature of the body at time t is T t) = T + T 0 T ) e kt, where T = 21 C and T 0 = 37 C. We denote t 0 = 0 the time when the victim died, t 1 the time in minutes that has elapsed since the body was recorded to be T t 1 ) = 31 C, and t 2 = 60 + t 1 to be the time when the body was discovered to be T t 2 ) = 29 C. We want to solve for t 1. First we solve for k. Consider the two equations T t 1 ) T = T 0 T ) e k t 1 T t 2 ) T = T 0 T ) e k t 2 = T t 1 ) T T t 2 ) T = e kt 2 t 1 ). Upon taking the logarithms of both sides, we can solve for the proportionality constant: Now we solve for t 1. We have k = 1 t 2 t 1 ln T t 1) T T t 2 ) T = 1 60 ln 31 21 29 21 = 0.00371906 minutes 1. T t 1 ) T = T 0 T ) e k t 1 = e k t 1 = T 0 T T t 1 ) T = t 1 = 1 k ln T 0 T T t 1 ) T. Hence t 1 = 60 ln / 37 21 31 21 ln = 126.377 minutes. 31 21 29 21 In other words, at midnight the victim had been dead for 126 minutes, so the victim died at 9:54 PM. 38
2.5 Radioactive Decay Problem 21. The half-life of a radioactive material is the time required for an amount of this material to decay to one-half its original value. Show that for any radioactive material that decays according to the equation dq = r Q, the half-life τ and the decay rate r satisfy the equation r τ = ln 2. Solution. First, we solve the differential equation: dq = r Q 1 dq = r Q 1 Q dq = r ln Q = r t + C 1 = Qt) = Q 0 e rt for some constant Q 0 = ±e C 1. The half-life τ is that time such that Qτ) = 0.5 Q 0. We have the equation Q 0 e rτ = 0.5 Q 0 so that e rτ = 2. Upon taking logarithms, we find that r τ = ln 2. Problem 22. Radium-266 has a half-life of 1620 years. Using the previous problem, find the time period during which a given amount of this material is reduced by one-quarter. Solution. Let Qt) denote the amount of some given initial amount Q 0 of Radium-266 after t years. We know that Qt) = Q 0 e rt for some decay rate r. The half-life is τ = 1620 yr, so we have r τ = ln 2, so that r = ln 2 = 0.000428 yr 1. We wish to find the time t such that τ Qt) = Q 0 1 4 Q 0 = 0.75 Q 0. We have the equation Q 0 e rt = 0.75 Q 0, so that e rt = 4 3. Hence t = ln4/3) = 672.361 yr. 0.000428 Problem 23. The isotope Iodine 131 is used to destroy tissue in an overactive thyroid gland. It has a half-life of 8.04 days. If a hospital receives a shipment of 500 mg of 131 I, how much of the isotope will be left after 20 days? Solution. Let Nt) denote the amount of 131 I; then Nt 0 ) = 500 mg when t 0 = 0 days. This amount satisfies the differential equation dn = r N for some decay rate r which we will find later. This equation is separable, so we find that Nt) = 500 e rt. Now we solve for the decay rate r. From the definition of half-life, when t 1 = 8.04 days we have Nt 1 ) = 1 2 N0), so 250 = 500 e 8.04 r. Taking 39
logarithms gives ln 250 = ln 500 8.04 r, so that r = ln 2 8.04 = 0.0862123 days 1. Hence the amount left after t 2 = 20 days is Nt 2 ) = 500 e r 20 = 500 e 0.0862123 20 = 5.60829 mg. Problem 24. Suppose that the ratio of 14 C to carbon in the charcoal on a cave wall is 0.617 times a similar ratio in living wood in the area. Use the Libby half-life to estimate the age of the charcoal. Solution. We use the formula found in the previous problem, where Rt) = R 0 e rt. We want to solve for t when Rt) = 0.617 R 0. We have 0.617 R 0 = R 0 e r t ln 0.617, so that t = = r 3878.99 years. Hence the age of the charcoal is approximately 3879 years. 40
Chapter 3 Exact Equations and Integrating Factors 3.1 Exact Equations Verify that the following equations are exact, and then find the solution. Problem 1. 2 x y 2 + 2 y ) + 2 x 2 y + 2 x ) dy = 0. Solution. Denote the functions P x, y) = 2 x y 2 + 2 y Qx, y) = 2 x 2 y + 2 x = P y = 4 x y + 2 Q x = 4 x y + 2 Hence the equation is indeed exact. We find a solution in the form F x, y) = C for the differential equation. Define the functions Gx, y) = P x, y) = x 2 y 2 + 2 x y, Hy) = [ Qx, y) G ] [2 dy = x 2 y + 2 x ) 2 x 2 y + 2 y )] dy = 0; y and set F x, y) = Gx, y) + Hy). Hence the solution is x 2 y 2 + 2 x y = C. Problem 2. x x 2 + y 2 ) 3/2 + y dy = 0. x 2 + y 2 3/2 ) Solution. Denote the functions P x, y) = x x 2 + y 2 ) 3/2 and Qx, y) = 41 y x 2 + y 2 ) 3/2.
We compute the partial derivatives P y = [ x x 2 + y 2) ] 3/2 = 3 y 2 x x 2 + y 2) 5/2 3 x y 2 y = x 2 + y 2 ) 5/2 ; Q x = [ y x 2 + y 2) ] 3/2 = 3 x 2 y x 2 + y 2) 5/2 3 x y 2 x = x 2 + y 2 ) 5/2. Hence the equation is indeed exact. We find a solution in the form F x, y) = C for the differential equation. Define the functions x Gx, y) = P x, y) = x 2 + y 2 ) 3/2 = 1 x 2 + y, 2 [ Hy) = Qx, y) G ] [ ] y dy = y x 2 + y 2 ) 3/2 y x 2 + y 2 ) 3/2 dy = 0; and set F x, y) = Gx, y) + Hy). Hence the solution is F x, y) = C 1 in terms of some constant C 1, which we can write as x 2 + y 2 = C for some constant C = C 1 2. Problem 3. x x 2 + y + y dy = 0. 2 x 2 + y2 Solution. Denote P x, y) = x y and Qx, y) = x 2 + y2. We have the partial derivative x 2 + y2 P y = y x x 2 + y 2) 1/2 x = x 2 + y 2) 3/2 x y 2 y = 2 x 2 + y 23 and so by symmetry P y = x y Q =. Hence the equation indeed is exact. It is easy x 2 + y23 x to verify that for F x, y) = x 2 + y 2 we have F F = P and = Q. Hence the solution to the x y differential equation is x 2 + y 2 = C for some constant C. Problem 4. Find the function y = yx) satisfying the differential equation 2 x y) +2 y x) dy = 0 as well as the initial condition y1) = 3. For what values x is this function valid? Solution. First we verify that this differential equation is exact. Denote the functions P x, y) = 2 x y Qx, y) = 2 y x 42 = P y = 1 Q x = 1
Now we compute F x, y) = Gx, y) + Hy) in terms of the functions Gx, y) = Hy) = P x, y) = x 2 x y, [ Qx, y) G ] dy = y [2 y x) x)] dy = 2 y dy = y 2. Hence F x, y) = Gx, y) + Hy) = x 2 x y + y 2, so that the desired solution y = yx) satisfies the implicit relation x 2 x y+y 2 = C for some constant C. Since y1) = 3, we see that C = F 1, 3) = 7. Hence yx) = 1 x + ) 28 3 x 2 2. This solution is valid when 28 3 x 2 0, which is the same 28 as saying x 3. Problem 5. Find the value of b for which the given equation is exact, and then solve it using that value of b. x y 2 + b x 2 y ) + x + y) x 2 dy = 0. Solution. Denote the functions P x, y) = x y 2 + b x 2 y Qx, y) = x + y) x 2 = x 3 + x 2 y = P y = 2 x y + b x2 Q x = 3 x2 + 2 x y The differential equation is exact when these derivatives are equal. They are equal when b = 3. We solve the differential by constructing a function F x, y) = Gx, y) + Hy). We define the functions Gx, y) = Hy) = P x, y) = 1 2 x2 y 2 + x 3 y, [ Qx, y) G ] [x dy = 3 + x 2 y ) x 2 y + x 3)] dy = 0. y The solution of the original differential equation is F x, y) = C, which we can write in the form x 2 y 2 + 2 x 3 y = C. 3.2 Equidimensional Equations Problem 6. Find the general solution of the equidimensional equation below. x 2 + y 2) 2 x y dy = 0. 43
Solution. We make the substitution y = t x to find a separable differential equation: x 2 + y 2) 2 x y dy = 0 dy = x2 + y 2 2 x y x + t = 1 + t2 2 t We find the general solution to this equation: = 1 x = 2 t x 1 t 2 2 t 1 t 2 = = ln x = ln 1 t 2 + C 1 = x = C 1 1 t 2 2 t x 1 t 2. for some constant C = ±e C 1. Substituting t = y x we find the equation x2 y 2 = C x. Problem 7. Find the general solution to the homogeneous equation below. x + y) + y x) dy = 0. Solution. We make the substitution y = t x to find a separable differential equation: x + y) + y x) dy = 0 dy = x + y x y x + t = 1 + t 1 t We find the general solution to this equation: = 1 t 1 + t 2 x 1 x = 1 t 1 + t 2 = ln x = arctan t 1 2 ln 1 + t 2 + C1 = t = tan = 1 t 1 + t 2 x. ln x 1 + t 2 C 1 ). This can be simplified a bit. Recall the formula tan A B) = tan A tan B 1 + tan A tan B = x tan ln ) x 2 + y 2 y y tan ln ) x 2 + y 2 + x = C where C = tan C 1 is a constant. 44
3.3 Bernoulli Equations Problem 8. Show that the change of variable, z = y 1 n, will transform the Bernoulli equation dy = P x) y + Qx) yn into the linear equation dz = 1 n) P x) z + 1 n) Qx). Here, we assume that n 1. Solution. If we substitute z = y 1 n, then we have dz Note that y = z y n. dy dy = 1 n) y n, and so = 1 dz yn 1 n. Hence the equation dy = P x) y + Qx) yn transforms to 1 n y n dz = P x) z y n + Qx) y n, so that dz = 1 n) P x) z + 1 n) Qx) as claimed. Use the technique of the previous exercise to transform the Bernoulli equations into a linear equation. Find the general solution of the resulting linear equation. Problem 9. dp = a P b P 2, assuming that a and b are constants. Solution. We set n = 2 and z = P 1. Then we have P = 1 z, P 2 = 1 dp, and z2 = 1 dz z 2. Hence the equation dp = a P b P 2 transforms to 1 dz z 2 = a z b, which simplifies to dz z2 + a z = b. In order to solve this equation, we multiply by the integrating factor µx) = e ax : d [ ] dz µx) z = µx) + dµ ) dz z = eax + a z = b e ax. Upon integrating we find µx) z = b a eax + C 1, and sozx) = b a + C 1 e at for some constant C 1. Hence P x) = 1 zx) = a b + C e ax for some constant C = C 1/a. Problem 10. x 2 dy + 2 x y y3 = 0. Solution. We sat n = 3 and z = y 2. Then we have y = z 1/2 and dy = 1 2 substitute these values into the differential equation: z 3/2 dz. We x 2 [ 1 2 ] dz z 3/2 + 2 x x 2 dy + 2 x y = y3 [ z 1/2] = x2 2 [z 1/2] 3 dz + 2 x z = 1 = dz 4 x z = 2 x 2. 45
This differential equation is linear, so we may solve it using integrating factors. We compute ) 4 µx) = exp x = exp 4 ln x ) = 1 x 4. Multiply the differential equation above by this function: 1 x 4 dz d dz 4 x z = 2 x 2 4 x 5 z = 2 x 6 [ ] 1 x 4 z = 2 x 6 = 1 x 4 z = 2 5 x 5 + C for some constant C. Hence, the solution to the differential equation is zx) = 2 + 5 C x5 1 5 x = yx) = = ± 5 x zx) 2 + 5 C x 5. 3.4 Integrating Factors Problem 11. The differential equation 3 y + 1) 2 x dy = 0 is not exact. Verify that µx, y) = y + 1 x 4 integrating factor, then solve the resulting exact equation. Solution. The resulting differential equation is 3 y + 1) 2 2 y + 1) x 4 x 3 dy = 0. First we verify that this equation is indeed an exact equation. Denote P x, y) = Qx, y) = 3 y + 1)2 x 4 2 y + 1) x 3 = We seek a function F x, y) = Gx, y) + Hy) such that Define the functions Gx, y) = Hy) = y + 1)2 Hence F x, y) = constant C. x 3 F x = P x, y) and F y y + 1)2 P x, y) = x 3, [ Qx, y) G ] dy = y P 6 y + 1) y x 4 Q x = Qx, y). [ 2 y + 1) x 3 + = 6 y + 1) x 4 ] 2 y + 1) x 3 dy = 0. = C, so the corresponding solution is y + 1) 2 + C x 3 = 0 for some 46