Solutions to Section PDF Free Download

Solutions to Section True-False Review: FALSE A derivative must involve some derivative of the function y f(x), not necessarily the first derivative TRUE The initial conditions accompanying a differential equation consist of the values of y, y, at t TRUE If we define positive velocity to be oriented downward, then where g is the acceleration due to gravity dv dt g, 4 TRUE We can justify this mathematically by starting from a(t) g, and integrating twice to get v(t) gt + c, and then s(t) gt + ct + d, which is a quadratic equation 5 FALSE The restoring force is directed in the direction opposite to the displacement from the equilibrium position 6 TRUE According to Newton s Law of Cooling, the rate of cooling is proportional to the difference between the object s temperature and the medium s temperature Since that difference is greater for the object at F than the object at 9 F, the object whose temperature is F has a greater rate of cooling 7 FALSE The temperature of the object is given by T (t) T m + ce kt, where T m is the temperature of the medium, and c and k are constants Since e kt, we see that T (t) T m for all times t The temperature of the object approaches the temperature of the surrounding medium, but never equals it 8 TRUE Since the temperature of the coffee is falling, the temperature difference between the coffee and the room is higher initially, during the first hour, than it is later, when the temperature of the coffee has already decreased 9 FALSE The slopes of the two curves are negative reciprocals of each other TRUE If the original family of parallel lines have slopes k for k, then the family of orthogonal trajectories are parallel lines with slope k If the original family of parallel lines are vertical (resp horizontal), then the family of orthogonal trajectories are horizontal (resp vertical) parallel lines FALSE The family of orthogonal trajectories for a family of circles centered at the origin is the family of lines passing through the origin Problems: d y dt g dy dt gt + c y(t) gt + c t + c Now impose the initial conditions y() c dy dt () c Hence, the solution to the initial-value problem is: y(t) gt The object hits the ground at time, t, when y(t ) Hence gt, so that t g 45 s, where we have taken g 98 ms From d y dt g, we integrate twice to obtain the general equations for the velocity and the position of the

ball, respectively: dy dt gt + c and y(t) gt + ct + d, where c, d are constants of integration Setting y to be at the top of the boy s head (and positive direction downward), we know that y() Since the object hits the ground 8 seconds later, we have that y(8) 5 (since the ground lies at the position y 5) From the values of y() and y(8), we find that d and 5 g + 8c Therefore, c 5 g 8 (a) The ball reaches its maximum height at the moment when y (t) That is, gt + c Therefore, t c g g 5 8g 98 s (b) To find the maximum height of the tennis ball, we compute y(98) 55 feet So the ball is 55 feet above the top of the boy s head, which is 585 feet above the ground From d y dt rocket, respectively: g, we integrate twice to obtain the general equations for the velocity and the position of the dy dt gt + c and y(t) gt + ct + d, where c, d are constants of integration Setting y to be at ground level, we know that y() Thus, d (a) The rocket reaches maximum height at the moment when y (t) That is, gt + c Therefore, the time that the rocket achieves its maximum height is t c At this time, y(t) 9 (the negative sign g accounts for the fact that the positive direction is chosen to be downward) Hence, ( 9 y c ) ( g g c ) ( + c c ) c g g g c g c g Solving this for c, we find that c ± 8g However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose c 8g 4 ms, and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 4 ms (b) The time that the rocket reaches its maximum height is t c g 4 98 4 From d y dt 48 s g, we integrate twice to obtain the general equations for the velocity and the position of the dy rocket, respectively: dt gt + c and y(t) gt + ct + d, where c, d are constants of integration Setting y to be at the level of the platform (with positive direction downward), we know that y() Thus, d (a) The rocket reaches maximum height at the moment when y (t) That is, gt + c Therefore, the time that the rocket achieves its maximum height is t c At this time, y(t) 85 (this is 85 m above g the platform, or 9 m above the ground) Hence, ( 85 y c ) ( g g c ) ( + c c ) c g g g c g c g

Solving this for c, we find that c ± 7g However, since c represents the initial velocity of the rocket, and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose c 7g 484 ms, and thus the initial speed at which the rocket must be launched for optimal viewing is approximately 484 ms (b) The time that the rocket reaches its maximum height is t c g 484 98 46 s 5 If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem We first integrate this differential equation: d y dt g, y(), dy () dt d y dt impose the initial conditions y() c dy g dt gt + c y(t) gt + c t + c Now dy dt () c Hence the soution to the intial-value problem is y(t) gt g() t We are given that y() h Consequently, h h (5g ) 47 m where we have taken g 98 ms 6 If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initial-value problem We first integrate the differential equation: d y dt the initial conditions y() c d y dt g, y(), dy dt () v dy g dt gt + c y(t) gt + c t + c Now impose dy dt () v c v Hence the soution to the intial-value problem is y(t) gt + v t We are given that y(t ) h Consequently, h gt + v t Solving for v yields v h gt t 7 y(t) A cos (ωt φ) dy dt Aω sin (ωt φ) d y dt Aω cos (ωt φ) + Aω cos (ωt φ) Aω cos (ωt φ) Hence, d y dt + ω y 8 y(t) c cos (ωt) + c sin (ωt) dy c ω sin (ωt) + c ω cos (ωt) d y dt dt c ω cos (ωt) c ω sin (ωt) ω c cos (ωt) + c cos (ωt) ω d y y Consequently, dt + ω y To determine the amplitude of the motion we write the solution to the differential equation in the equivalent form: y(t) c + c We can now define an angle φ by c cos (ωt) + c + c c sin (ωt) c + c cos φ c c + c and sin φ c c + c

4 Then the expression for the solution to the differential equation is y(t) c + c cos (ωt) cos φ + sin (ωt) sin φ c + c cos (ωt + φ) Consequently the motion corresponds to an oscillation with amplitude A c + c 9 We compute the first three derivatives of y(t) ln t: Therefore, as required dy dt t, d y dt t, d y dt t ( ) dy dt t d y dt, We compute the first two derivatives of y(x) x/(x + ): dy dx (x + ) and d y dx (x + ) Then y + d y dx as required x x + (x + ) x + x + x (x + ) (x + ) + (x + x ) (x + ) We compute the first two derivatives of y(x) e x sin x: (x + ) + x + x ( + x), Then as required dy dx ex (sin x + cos x) and d y dx ex cos x y cot x d y dx (ex sin x) cot x e x cos x, dt (T T m ) dt k d dt (ln T T m ) k The preceding equation can be integated directly to yield ln T T m kt + c Exponentiating both sides of this equation gives T T m e kt+c, which can be written as T T m ce kt, where c ±e c Rearranging yields T (t) T m + ce kt After 4 pm In the first two hours after noon, the water temperature increased from 5 F to 55 F, an increase of five degrees Because the temperature of the water has grown closer to the ambient air temperature, the temperature difference T T m is smaller, and thus, the rate of change of the temperature of the water grows smaller, according to Newton s Law of Cooling Thus, it will take longer for the water temperature to increase another five degrees Therefore, the water temperature will reach 6 F more than two hours later than pm, or after 4 pm

5 4 The object temperature cools a total of 4 F during the 4 minutes, but according to Newton s Law of Cooling, it cools faster in the beginning (since T T m is greater at first) Thus, the object cooled half-way from 7 F to F in less than half the total cooling time Therefore, it took less than minutes for the object to reach 5 F 5 Given a family of curves satisfies: x + 4y c x + 8y dy dx Orthogonal trajectories satisfy: dy dx x 4y dy dx 4y x dy y dx 4 x d dx (ln y ) 4 x ln y 4 ln x + c y kx 4, where k ±e c y(x) 8 4 x -5 - -5-4 5 5-8 Figure : Figure for Exercise 5 6 Given a family of curves satisfies: y c x Orthogonal trajectories satisfy: dy dx x y y dy dx x d dx dy dy x + y dx dx y x ( ) y x y x + c y x c, where c c 7 Given family of curves satisfies: y cx c y x Hence, Orthogonal trajectories satisfy: dy ( y ) dx cx c x x y x dy dx x dy d y x y dx dx (y ) x y x + c y + x c, where c c 8 Given family of curves satisfies: y cx 4 c y x 4 Hence, dy ( y ) dx 4cx 4 x 4 x 4y x

6 y(x) 4 x -4-4 - -4 Figure : Figure for Exercise 6 y(x) 6 8 4 x - - -4-8 - -6 - Figure : Figure for Exercise 7 Orthogonal trajectories satisfy: where c c dy dx x dy d 4y x 4y dx dx (y ) x y x + c 4y + x c,

7 y(x) 8 4 x -5-5 -4-8 Figure 4: Figure for Exercise 8 9 Given family of curves satisfies: y x + c dy dx Orthogonal trajectories satisfy: y dy dy d y y dx dx dx (ln y ) ln y x + c y c e x 4 y(x) x - 4 - - - -4 Figure 5: Figure for Exercise 9 Given family of curves satisfies: y ce x dy dx cex y Orthogonal trajectories satisfy: dy dx ( ) dy d y y dx dx y y x + c y x + c

8 y(x) - x - - Figure 6: Figure for Exercise y mx + c dy dx m Orthogonal trajectories satisfy: dy dx m y m x + c y cx m dy dx cmxm, but c dy dx x dy y my dx x m d dx y + mx c y dy dy + mx dx dx mx y Orthogonal trajectories satisfy: y dy so xm dx my Orthogonal trajectories satisfy: x ( ) y x m y m x + c y m x + c dy dx y dy y mx dx mx d (ln y ) dx mx m ln y ln x + c y m c x 4 y mx + c y dy dy m dx dx m y Orthogonal trajectories satisfy: dy dx y m y dy dx m d dx (ln y ) m ln y m x + c y c e x m 5 u x + y x + 4y dy dy dx dx x y

9 Orthogonal trajectories satisfy: dy dx y x y dy dx x d dx (ln y ) x ln y ln x + c y c x y(x) 6 8 4 x - - -4-8 - -6 - Figure 7: Figure for Exercise 5 6 m tan (a ) tan (a a) tan (a ) tan (a) + tan (a ) tan (a) m tan (a) + m tan (a) True-False Review: Solutions to Section FALSE The order of a differential equation is the order of the highest derivative appearing in the differential equation TRUE This is condition in Definition TRUE This is the content of Theorem 5 4 FALSE There are solutions to y + y that do not have the form c cos x + 5c cos x, such as y(x) sin x Therefore, c cos x + 5c cos x does not meet the second requirement set forth in Definition for the general solution 5 FALSE There are solutions to y + y that do not have the form c cos x + 5c sin x, such as y(x) cos x + sin x Therefore, c cos x + 5c sin x does not meet the second requirement set form in Definition for the general solution 6 TRUE Since the right-hand side of the differential equation is a function of x only, we can integrate both sides n times to obtain the formula for the solution y(x)

Problems:, nonlinear, linear, nonlinear 4, nonlinear 5 4, linear 6, nonlinear 7 We can quickly compute the first two derivatives of y(x): y (x) (c +c )e x cos x+( c +c )e x sin x and y (x) ( c +4c )e x cos x+( 4c c )e x sin x Then we have y y + 5y ( c + 4c )e x cos x + ( 4c c )e x sin x (c + c )e x cos x + ( c + c )e x sin x+5(c e x cos x+c e x sin x), which cancels to, as required This solution is valid for all x R 8 y(x) c e x + c e x y c e x c e x y c e x + 4c e x y + y y (c e x + 4c e x ) + (c e x c e x ) (c e x + c e x ) Thus y(x) c e x + c e x is a solution of the given differential equation for all x R 9 y(x) x + 4 y (x + 4) y Thus y(x) is a solution of the given differential x + 4 equation for x (, 4) or x ( 4, ) y(x) c x y c x y x Thus y(x) c x is a solution of the given differential equation for all x {x : x > } y(x) c e x sin (x) y c e x cos (x) c e x sin (x) y c e x sin (x) 4c e x cos (x) y +y +5y c e x sin (x) 4c e x cos (x)+c e x cos (x) c e x sin (x)+5c e x sin (x) Thus y(x) c e x sin (x) is a solution to the given differential equation for all x R y(x) c cosh (x) + c sinh (x) y c sinh (x) + c cosh (x) y 9c cosh (x) + 9c sinh (x) y 9y 9c cosh (x) + 9c sinh (x) 9c cosh (x) + c sinh (x) Thus y(x) c cosh (x) + c sinh (x) is a solution to the given differential equation for all x R y(x) c x + c x y c x 4 c x y c x 5 + c ( x x y + 5xy + y x c x 5 + c ) x + ( 5x c x 4 c ) ( c x + x + c ) Thus y(x) c x x + c is a solution to the given differential equation x for all x (, ) or x (, ) 4 y(x) c x+x y c x +6x y c ( 4 x +6 x y xy +y x c ) 4 x + 6

( ) c x x + 6x +(c x+x ) 9x Thus y(x) c x+x is a solution to the given differential equation for all x {x : x > } 5 y(x) c x + c x x sin x y c x + c x x cos x x sin x y c + 6c x + x sin x x cos x x cos sin x Substituting these results into the given differential equation yields x y 4xy + 6y x (c + 6c x + x sin x 4x cos x sin x) 4x(c x + c x x cos x x sin x) + 6(c x + c x x sin x) c x + 6c x + x 4 sin x 4x cos x x sin x 8c x c x + 4x cos x + 8x sin x x 4 sin x Hence, y(x) c x + c x x sin x is a solution to the differential equation for all x R + 6c x + 6c x 6x sin x 6 y(x) c e ax + c e bx y ac e ax + bc e bx y a c e ax + b c e bx Substituting these results into the differential equation yields y (a + b)y + aby a c e ax + b c e bx (a + b)(ac e ax + bc e bx ) + ab(c e ax + c e bx ) (a c a c abc + abc )e ax + (b c abc b c + abc )e bx Hence, y(x) c e ax + c e bx is a solution to the given differential equation for all x R 7 y(x) e ax (c + c x) y e ax (c ) + ae ax (c + c x) e ax (c + ac + ac x) y ea ax (ac ) + ae ax (c + ac + ac x) ae ax (c + ac + ac x) Substituting these into the differential equation yields y ay + a y ae ax (c + ac + ac x) ae ax (c + ac + ac x) + a e ax (c + c x) ae ax (c + ac + ac x c ac ac x + ac + ac x) Thus, y(x) e ax (c + c x) is a solution to the given differential eqaution for all x R 8 y(x) e ax (c cos bx + c sin bx) so, y e ax ( bc sin bx + bc cos bx) + ae ax (c cos bx + c sin bx) e ax (bc + ac ) cos bx + (ac bc ) sin bx so, y e ax b(bc + ac ) sin bx + b(ac + bc ) cos bx + ae ax (bc + ac ) cos bx + (ac + bc ) sin bx e ax (a c b c + abc ) cos bx + (a c b c abc ) sin bx Substituting these results into the differential equation yields y ay + (a + b )y (e ax (a c b c + abc ) cos bx + (a c b c abc ) sin bx) a(e ax (bc + ac ) cos bx + (ac bc ) sin bx) + (a + b )(e ax (c cos bx + c sin bx)) e ax (a c b c + abc abc a c + a c + b c ) cos bx + (a c b c abc + abc a c + a c + b c ) sin bx Thus, y(x) e ax (c cos bx + c sin bx) is a solution to the given differential equation for all x R 9 y(x) e rx y re rx y r e rx Substituting these results into the given differential equation yields e rx (r + r ), so that r must satisfy r + r, or (r + )(r ) Consequently r and r are the only values of r for which y(x) e rx is a solution to the given differential equation The corresponding solutions are y(x) e x and y(x) e x

y(x) e rx y re rx y r e rx Substitution into the given differential equation yields e rx (r 8r + 6), so that r must satisfy r 8r + 6, or (r 4) Consequently the only value of r for which y(x) e rx is a solution to the differential equation is r 4 The corresponding solution is y(x) e 4x y(x) x r y rx r y r(r )x r Substitution into the given differential equation yields x r r(r ) + r, so that r must satisfy r Consequently r and r are the only values of r for which y(x) x r is a solution to the given differential equation The corresponding solutions are y(x) x and y(x) x y(x) x r y rx r y r(r )x r Substitution into the given differential equation yields x r r(r ) + 5r + 4, so that r must satisfy r + 4r + 4, or equivalently (r + ) Consequently r is the only value of r for which y(x) x r is a solution to the given differential equation The corresponding solution is y(x) x y(x) x(5x ) (5x x) y (5x ) y 5x Substitution into the Legendre equation with N yields ( x )y xy + y ( x )(5x) + x(5x ) + 6x(5x ) Consequently the given function is a solution to the Legendre equation with N 4 y(x) a +a x+a x y a +a x y 4a Substitution into the given differential equation yields ( x )(a ) x(a +a x)+4(a +a x+a x ) a x+a +4a For this equation to hold for all x we require a, and a + 4a Consequently a, and a a The corresponding solution to the differential equation is y(x) a ( x ) Imposing the normalization condition y() requires that a Hence, the required solution to the differential equation is y(x) x 5 x sin y e x c x cos y dy dx + sin y ex dy dx ex sin y x cos y 6 xy + y x c xy dy dx + y + dy dy dx dx y (xy + ) 7 e xy + x c e xy x dy dy + y xexy dx dx + yexy yexy xe xy Given y() e () c c Therefore, e xy x, so that y ln x x 8 e y/x + xy x c e y/x x dy dx y x + xy dy dx + y dy dx x ( y ) + ye y/x x(e y/x + x y) 9 x y sin x c x y dy dx + xy cos x dy cos x xy dx x Since y(π) y π, then ( ) π sin π c c Hence, x y sin x so that y + sin x π x Since y(π), take the π + sin x branch of y where x < so y(x) x dy sin x y(x) cos x + c for all x R dx

dy dx x / y(x) x / + c for all x > d y dx xex dy dx xex e x + c y(x) xe x e x + c x + c for all x R d y dx xn, where n is an integer If n then dy dx ln x + c y(x) x ln x + c x + c for all x (, ) or x (, ) If n then dy dx x + c y(x) c x + c ln x for all x (, ) or x (, ) If n and n then dy dx xn+ n + + c x n+ (n + )(n + ) + c x + c for all x R dy 4 ln x y(x) x ln x x + c Since, y() () + c c Thus, dx y(x) x ln x x + 5 d y dy cos x dx dx sin x + c y(x) cos x + c x + c Thus, y () c, and y() c Thus, y(x) + x cos x 6 d y dx 6x d y dx x + c dy dx x + c x + c y 4 x4 + c x + c x + c Thus, y () 4 c 4, and y () c, and y() c Thus, y(x) 4 x4 + x x + 7 y xe x y xe x e x + c y xe x e x + c x + c Thus, y () 4 c 5, and y() c 5 Thus, y(x) xe x e x + 5x + 5 8 Starting with y(x) c e x + c e x, we find that y (x) c e x c e x and y (x) c e x + c e x Thus, y y, so y(x) c e x + c e x is a solution to the differential equation on (, ) Next we establish that every solution to the differential equation has the form c e x + c e x Suppose that y f(x) is any solution to the differential equation Then according to Theorem 5, y f(x) is the unique solution to the initial-value problem y y, y() f(), y () f () However, consider the function y(x) f() + f () e x + f() f () e x This is of the form y(x) c e x + c e x, where c f()+f () and c f() f (), and therefore solves the differential equation y y Furthermore, evaluation this function at x yields y() f() and y () f () Consequently, this function solves the initial-value problem above However, by assumption, y(x) f(x) solves the same initial-value problem Owing to the uniqueness of the solution to this initial-value problem, it follows that these two solutions are the same: f(x) c e x + c e x

4 Consequently, every solution to the differential equation has the form y(x) c e x + c e x, and therefore this is the general solution on any interval I d y 9 dx e x dy dx e x + c y(x) e x + c x + c Thus, y() c, and y() c e Hence, y(x) e x e x 4 d y dy 6 4 ln x dx dx x 4x ln x + c y(x) x ln x + c x + c Since, y() c + c, and since, y(e) ec + c e Solving this system yields c Thus, y(x) e e (x ) x ln x e e, c e e 4 y(x) c cos x + c sin x (a) y() c () + c () c y(π) c (), which is impossible No solutions (b) y() c () + c () c y(π) c (), so c can be anything Infinitely many solutions 4-47 Use some kind of technology to define each of the given functions Then use the technology to simplify the expression given on the left-hand side of each differential equation and verify that the result corresponds to the expression on the right-hand side 48 (a) Use some form of technology to substitute y(x) a + bx + cx + dx + ex 4 + fx 5 where a, b, c, d, e, f are constants, into the given Legendre equation and set the coefficients of each power of x in the resulting equation to zero The result is: e, f + 8d, e + c, d + 4b, c + 5a Now solve for the constants to find: a c e, d 4 b, f 9 d 5 b Consequently the corresponding solution to the Legendre equation is: ( y(x) bx 4 x + ) 5 x4 Imposing the normalization condition y() requires b( 4 required solution is y(x) 8 x(5 7x + 6x 4 ) + 5 ) b 5 8 Consequently the 49 (a) J (x) k ( ) k ( x ) k (k!) 4 x + 64 x4 + (b) A Maple plot of J(, x, 4) is given in the acompanying figure From this graph, an approximation to the first positive zero of J (x) is 4 Using the Maple internal function BesselJZeros gives the approximation 4485558 (c) A Maple plot of the functions J (x) and J(, x, 4) on the interval, is given in the acompanying figure We see that to the printer resolution, these graphs are indistinguishable On a larger interval, for example,,, the two gaphs would begin to differ dramatically from one another (d) By trial and error, we find the smallest value of m to be m A plot of the functions J(, x) and J(, x, ) is given in the acompanying figure

5 J(, x, 4) 8 6 4 Approximation to the first positive zero of J (x) 4 x Figure 8: Figure for Exercise 49(b) J (x), J(, x, 4) 8 6 4 4 6 8 4 6 8 x Figure 9: Figure for Exercise 49(c) Solutions to Section True-False Review: TRUE This is precisely the remark after Theorem FALSE For instance, the differential equation in Example 6 has no equilibrium solutions FALSE This differential equation has equilibrium solutions y(x) and y(x) 4 TRUE For this differential equation, we have f(x, y) x + y Therefore, any equation of the form x + y k is an isocline, by definition 5 TRUE Equilibrium solutions are always horizontal lines These are always parallel to each other 6 TRUE The isoclines have the form x +y y k, or x + y ky, or x + (y k) k, so the statement is valid

6 J(, x), J(, x, ) 8 6 4 4 4 6 8 J(, x, ) x J(, x) Figure : Figure for Exercise 49(d) 7 TRUE An equilibrium solution is a solution, and two solution curves to the differential equation f(x, y) do not intersect dy dx Problems: y cx c xy Hence, dy dx cx (xy)x y x y cx c y dy Hence, x dx cx y x x y x x + y cx x + y x Consequently, dy dx y x xy c Hence, x + y dy dx c x + y x, so that, y dy dx x + y x x 4 y cx c y dy dy Hence, y c, so that, x dx dx c y y x 5 cy x c c + cy x c y ± 4y + 4x y ± x + y 6 y x c y dy dy x dx dx x y 7 (x c) + (y c) c x cx + y cy c x + y Differentiating the given (x + y) equation yields (x c) + (y c) dy, so that x x + y + y x + y dy, that is dx (x + y) (x + y) dx dy dx + xy y x y + xy x

7 8 x + y c x + y dy dy dx dx x y y(x) 6 8 4 - - -4 x -8 - -6 - Figure : Figure for Exercise 8 9 y cx dy dx cx y x x y x The initial condition y() 8 8 c() c Thus the unique solution to the initial value problem is y x y(x) 8 (, 8) 4 - x -4-8 Figure : Figure for Exercise 9

8 y cx y dy dy c y dx dx y dy yx x dy y dx The initial condition x dx y() c 4, so that the unique solution to the initial value problem is y 4x y(x) (, ) - - - - x - Figure : Figure for Exercise (x c) + y c x cx + c + y c, so that x cx + y () Differentiating with respect to x yields x c + y dy () dx But from (), c x + y ( x + y ) which, when substituted into (), yields x + y dy x x dx, that is, dy dx y x Imposing the initial condition y() from () c, so that the unique xy solution to the initial value problem is y + x(4 x) Let f(x, y) x sin (x + y), which is continuous for all x, y R f x cos (x + y), which is continuous for all x, y R y By Theorem, dy dx x sin (x + y), y(x ) y has a unique solution for some interval I R dy dx x x + (y 9), y() f(x, y) x x + (y 9), which is continuous for all x, y R f y xy x, which is continuous for all x, y R +

9 y(x) (, ) 4 5 6 x - - - Figure 4: Figure for Exercise So the IVP stated above has a unique solution on any interval containing (, ) By inspection we see that y(x) is the unique solution 4 The IVP does not necessarily have a unique solution since the hypothesis of the existence and uniqueness theorem are not satisfied at (,) This follows since f(x, y) xy /, so that f y xy / which is not continuous at (, ) 5 (a) f(x, y) xy f 4xy Both of these functions are continuous for all (x, y), and therefore y the hypothesis of the uniqueness and existence theorem are satisfied for any (x, y ) (b) y(x) x + c y x (x + c) xy (c) y(x) x + c (i) y() c Hence, y(x) c x The solution is valid on the interval (, ) + (ii) y() + c c Hence, y(x) This solution is valid on the interval (, ) x (iii) y() c c Hence, y(x) x This solution is valid on the interval (, ) (d) Since, by inspection, y(x) satisfies the given IVP, it must be the unique solution to the IVP 6 (a) Both f(x, y) y(y ) and f y are continuous at all points (x, y) Consequently, the y hypothesis of the existence and uniqueness theorem are satisfied by the given IVP for any x, y (b) Equilibrium solutions: y(x), y(x)

y(x) 8 4 - x Figure 5: Figure for Exercise 5c(i) 8 y(x) 6 4 x 4 5 6 Figure 6: Figure for Exercise 5c(ii) (c) Differentiating the given differential equation yields d y dy dx (y ) (y )y(y ) Hence the dx solution curves are concave up for < y <, and y >, and concave down for y <, and < y < (d) The solutions will be bounded provided y 7 y 4x There are no equalibrium solutions The slope of the solution curves is positive for x > and is negative for x < The isoclines are the lines x k 4 Slope of Solution Curve Equation of Isocline -4 x - x / x x / 4 x

y(x) - -5 5 x - - - Figure 7: Figure for Exercise 5c(iii) y(x) x Figure 8: Figure for Exercise 6(d) 8 y x There are no equalibrium solutions The slope of the solution curves is positive for x > and increases without bound as x + The slope of the curve is negative for x < and decreases without bound as x The isoclines are the lines x k

y(x) 5 5 5 5 5 5 5 x 5 Figure 9: Figure for Exercise 7 Slope of Solution Curve Equation of Isocline ±4 x ±/4 ± x ±/ ±/ x ± ±/4 x ±4 ±/ x ± y(x) x Figure : Figure for Exercise 8 9 y x + y There are no equalibrium solutions The slope of the solution curves is positive for y > x, and negative for y < x The isoclines are the lines y + x k Slope of Solution Curve Equation of Isocline y x y x y x y x + y x +

Since the slope of the solution curve along the isocline y x coincides with the slope of the isocline, it follows that y x is a solution to the differential equation Differentiating the given differential equation yields: y + y + x + y Hence the solution curves are concave up for y > x, and concave down for y < x Putting this information together leads to the slope field in the acompanying figure y(x) - - - x - - - Figure : Figure for Exercise 9 y x y There are no equalibrium solutions The slope of the solution curves is zero when x The solution has a vertical tangent line at all points along the x-axis (except the origin) Differentiating the differential equation yields: y y x y y y x y y (y x ) Hence the solution curves are concave up for y > and y > x ; y < and y < x and concave down for y > and y < x ; y < and y > x The isoclines are the lines x y k Slope of Solution Curve Equation of Isocline ± y ±x/ ± y ±x ±/ y ±x ±/4 y ±4x ±/ y ±x Note that y ±x are solutions to the differential equation y 4x y Slope is zero when x (y ) The solutions have a vertical tangent line at all points along the x-axis(except the origin) The isoclines are the lines 4x y k Some values are given in the table below

4 y(x) x Figure : Figure for Exercise Slope of Solution Curve Equation of Isocline ± y ±4x ± y ±x ± y ±4x/ Differentiating the given differential equation yields: y 4 y + 4xy y 4 y 6x y 4(y + 4x ) y Consequently the solution curves are concave up for y <, and concave down for y > Putting this information together leads to the slope field in the acompanying figure y x y Equalibrium solution: y(x) no solution curve can cross the x-axis Slope: zero when x or y Positive when y > (x ), negative when y < (x ) Differentiating the given differential equation yields: d y dy xy+x dx dx xy+x4 y xy(+x ) So, when y >, the solution curves are concave up for x (, ( ) / ), and for x >, and are concave down for x (( ) /, ) When y <, the solution curves are concave up for x (( ) /, ), and concave down for x (, ( ) / ) and for x > The isoclines are the hyperbolas x y k Slope of Solution Curve Equation of Isocline ± y ±/x ± y ±/x ±/ y ±/(x) ±/4 y ±/(4x) ±/ y ±/(x) y y x cos y The slope is zero when x There are equalibrium solutions when y (k + ) π The slope field is best sketched using technology The accompanying figure gives the slope field for π < y < π 4 y x + y The slope of the solution curves is zero at the origin, and positive at all the other points There are no equilibrium solutions The isoclines are the circles x + y k

5 y(x) 4 - - x Figure : Figure for Exercise y(x) x Figure 4: Figure for Exercise Slope of Solution Curve Equation of Isocline x ±/4 x ±/ x ± 4 x ±4 5 x ±

6 y(x) 5 4 x Figure 5: Figure for Exercise y(x) x Figure 6: Figure for Exercise 4 dt 5 dt 8 (T 7) Equilibrium solution: T (t) 7 The slope of the solution curves is positive for d T T > 7, and negative for T < 7 dt dt 8 dt (T 7) Hence the solution curves are concave 64 up for T > 7, and concave down for T < 7 The isoclines are the horizontal lines 8 (T 7) k Slope of Solution Curve Equation of Isocline /4 T 9 /5 T 86 T 7 /5 T 54 /4 T 5 6 y xy 7 y x sin x + y

7 T(t) 8 6 4 4 6 8 4 t Figure 7: Figure for Exercise 5 y(x) x Figure 8: Figure for Exercise 6 y(x) x Figure 9: Figure for Exercise 7 8 y x y

8 y(x) x Figure : Figure for Exercise 8 9 y x sin y y(x) x Figure : Figure for Exercise 9 y + y + 5x y y + 5x (a) Slope field for the differential equation y x ( sin x y) (b) Slope field with solution curves included The figure suggests that the solution to the differential equation are unbounded as x + (c) Slope field with solution curve corresponding to the initial condition y( π ) 6 π This solution curve is bounded as x + (d) In the accompanying figure we have sketched several solution curves on the interval (,5 The figure suggests that the solution curves approach the x-axis as x

9 y(x) x Figure : Figure for Exercise y(x) x Figure : Figure for Exercise y(x) 4 4 6 8 x 4 Figure 4: Figure for Exercise (a) (a) Differentiating the given equation gives dy dx kx y x Hence the differential equation of the

y(x) 4 4 6 8 x 4 Figure 5: Figure for Exercise (b) y(x) 4 4 6 8 x 4 Figure 6: Figure for Exercise (c) y(x) 4 4 6 8 4 x 4 Figure 7: Figure for Exercise (d) othogonal trajectories is dy dx x y

y(x) 4 4 4 x 4 Figure 8: Figure for Exercise (a) (b) The orthogonal trajectories appear to be ellipses This can be verified by intergrating the differential equation derived in (a) 4 If a >, then as illustrated in the following slope field (a 5, b ), it appears that lim t i(t) b a y(x) 4 x 4 Figure 9: Figure for Exercise 4 when a > If a <, then as illustrated in the following slope field (a 5, b ) it appears that i(t) diverges as t If a and b, then once more i(t) diverges as t The acompanying figure shows a representative case when b > Here we see that lim t i(t) + If b <, then lim t i(t) If a b, then the general solution to the differential equation is i(t) i where i is a constant Solutions to Section 4

4 i(t) t 4 Figure 4: Figure for Exercise 4 when a < 4 y(x) x 4 Figure 4: Figure for Exercise 4 when a True-False Review: TRUE The differential equation dy dx f(x)g(y) can be written according to Definition 4, for a separable differential equation dy g(y) dx f(x), which is the proper form, TRUE A separable differential equation is a first-order differential equation, so the general solution contains one constant The value of that constant can be determined from an initial condition, as usual TRUE Newton s Law of Cooling is usually expressed as dt dt and this form shows that the equation is separable dt T T m dt k, k(t T m), and this can be rewritten as 4 FALSE The expression x + y cannot be separated in the form f(x)g(y), so the equation is not separable 5 FALSE The expression x sin(xy) cannot be separated in the form f(x)g(y), so the equation is not separable

y dy 6 TRUE We can write the given equation as e dx equation ex, which is the proper form for a separable 7 TRUE We can write the given equation as ( + y ) dy dx x, which is the proper form for a separable equation 8 FALSE The expression x+4y 4x+y cannot be separated in the form f(x)g(y), so the equation is not separable 9 TRUE We can write x y+x y x +xy xy, so we can write the given differential equation as dy y dx x, which is the proper form for a separable equation Problems: Separating the variables and integrating yields dy y xdx ln y x + c y(x) ce x Separating the variables and integrating yields y dx dy x + y(x) tan x + c Separating the variables and integrating yields e y dy e x dx e y + e x c y(x) ln (c e x ) 4 Separating the variables and integrating yields dy (ln x) y dx y(x) c ln x x 5 Separating the variables and integrating yields dx dy x y ln x ln y c y(x) c(x ) 6 Separating the variables and integrating yields dy y x x + dx ln y ln x + + c y(x) c(x + ) + 7 y x dy dy x dx dx dy x(x ) ( y) Separating the variables and integrating yields dx dx dx ln y x(x ) x + x dx dy y ln y ln x + ln x + c x (y )(x ) c y(x) cx x 8 dy cos (x y) dy cos x cos y dx sin x sin y dx sin x sin y sin y cos y dy cos x cos y dx ln cos y ln sin x +c cos y c csc x

4 9 dy dx x(y ) (x )(x ) dy (y + )(y ) dy y + + dy y ( dx x xdx, y ± Thus, (x )(x ) ) dx ln y + +ln y ln x ln x +c x y ) (x ) + c(x ) c(x y(x) By inspection we see that y(x), and y(x) y + x (x ) c(x ) are solutions of the given differential equation The former is included in the above solution when c dy dx x y 6 x + dy y x 6 x x 6 ( ln y x 6 8 dx x 6 ln x 4 + c y(x) + c ( x + 4 x 4 ) e x dx ln y dx x + 4 + 8 ( + 6 ) x dx ln y 6 ) dx ln y x + ln x + 4 x 4 (x a)(x b) dy dy (y c) dx y c dx (x a)(x b) dy y c ( a b x a ) dx x b x a /(a b) ( ) /(a b) ln y c ln c x b ( ) /(a b) x a (y c) x b x a c y c c x b ( ) /(a b) x a y(x) c + c x b (x + ) dy dx + y Thus, tan y tan x + π 4 dy + y dx + x tan y tan x + c, but y() so c π 4 or y(x) x + x ( x ) dy dy + xy ax dx a y x x dx ln a y ln x + c y(x) a + c x, but y() a so c a and therefore, y(x) a( + x ) 4 dy sin (x + y) dy dx sin x sin y dx tan x cot y sin y cos y dy sin x dx ln cos x cos y c, but cos x y( π 4 ) π 4 so c ln () Hence, ln cos x cos y ln () y(x) ( cos sec x) 5 dy dx y sin x dy y sin xdx for y Thus cos x+c However, we cannot impose the y initial condition y() on the last equation since it is not defined at y But, by inspection, y(x) is a solution to the given differential equation and further, y() ; thus, the unique solution to the initial value problem is y(x) dy 6 dx (y )/ dy (y ) / dx if y (y ) / x + c but y() so c y x y (x ) This does not contradict the Existance-Uniqueness theorem because the hypothesis of the theorem is not satisfied when x

7(a) m dv dt mg kv be written as m k m k (mg/k) v dv dt If we let a mg k a v dv dt which can be integrated directly to obtain that is, upon exponentiating both sides, ( ) m a + v ak ln t + c, a v a + v a v c e ak m t Imposing the initial condition v(), yields c so that a + v a v e ak m t 5 then the preceding equation can Therefore, ( ) e akt m v(t) a e akt m + which can be written in the equivalent form v(t) a tanh ( ) gt a (b) No As t, v a and as t +, v (c) v(t) a tanh ( ) gt dy a dt a tanh ( ) ( gt a a tanh gt ) a a dt y(t) g y() then y(t) a g ln cosh ( gt a ) ln(cosh ( gt a )) + c and if 8 The required curve is the solution curve to the IVP dy dx x 4y, y() Separating the variables in the differential equation yields 4y dy dx, which can be integrated directly to obtain y x + c Imposing the initial condition we obtain c, so that the solution curve has the equation y x +, or equivalently, 4y + x 9 The required curve is the solution curve to the IVP dy dx ex y, y() Separating the variables in the differential equation yields e y dy e x dx, which can be integrated directly to obtain e y e x + c Imposing the initial condition we obtain c e e, so that the solution curve has the equation e y e x + e e, or equivalently, y ln(e x + e e ) The required curve is the solution curve to the IVP dy dx x y, y( ) Separating the variables in the differential equation yields y dy x dx, which can be integrated directly to obtain y x + c Imposing the initial condition we obtain c or equivalently, y x, so that the solution curve has the equation y x,

6 (a) Separating the variables in the given differential equation yields dv dt Integrating we + v obtain tan (v) t + c The initial condition v() v implies that c tan (v ), so that tan (v) t+tan (v ) The object will come to rest if there is time t, at which the velocity is zero To determine t r, we set v in the previous equation which yields tan () t r +tan (v ) Consequently, t r tan (v ) The object does not remain at rest since we see from the given differential equation that dv dt < att t r, and so v is decreasing with time Consequently v passes through zero and becomes negative for t < t r (b) From the chain rule we have dv dt dx dv Then dt dx v dv Substituting this result into the differential dx equation (47) yields v dv dx ( + v v ) We now separate the variables: dv dx Integrating we + v obtain ln ( + v ) x + c Imposing the initial condition v() v, x() implies that c ln ( + v), so that ln ( + v ) x + ln ( + v) When the object comes to rest the distance traveled by the object is x ln ( + v ) (a) dv dt kvn v n dv kdt n : n v n kt + c Imposing the initial condition v() + v yields c n v n, so that v v n + (n )kt /( n) The object comes to rest in a finite time if there is a positive value of t for which v n : Integratingv n dv kdt and imposing the initial conditions yields v v e kt, and the object does not come to rest in a finite amount of time (b) If n,, then dx dt v n + (n )kt /( n), where x(t) denotes the distanced traveled by the object Consequently, x(t) k( n) v n + (n )kt ( n)/( n) + c Imposing the initial condition x() yields c k( n) v n, so that x(t) k( n) v n o + n(n )kt ( n)/( n) + k( n) v n For < n <, we have n n <, so that lim t x(t) Hence the maximum distance that the k( n) object can travel in a finite time is less than k( n) If n, then we can integrate to obtain x(t) v k ( e kt ), where we have imposed the initial condition x() Consequently, lim t x(t) v Thus in this case the maximum distance that the object can k travel in a finite time is less than v k (c) If n >, then x(t) k( n) v n o + n(n )kt ( n)/( n) + k( n) v n is still valid However, in this case n n >, and so lim t x(t) + COnsequently, there is no limit to the distance that the object can travel If n, then we return to v v n + (n )kt /( n) In this case dx dt (v + kt), which can be integrated directly to obtain x(t) k ln ( + v kt), where we have imposed the initial condition that x() Once more we see that lim t x(t) +, so that there is no limit to the distance that the object can travel

7 Solving p p ( ρ ρ ) /γ Consequently the given differential equation can be written as dp gρ ( p p ) /γ dy, or equivalently, p /γ dp gρ dy This can be integrated directly to obtain γp(γ )/γ gρ y + c At p /γ γ p /γ the center of the Earth we have p p Imposing this initial condition on the preceding solution gives c γp(γ )/γ γ Substituting this value of c into the general solution to the differential equation we find, (γ )ρ gy, so that p p (γ )ρ (γ )/γ gy γp γp after some simplification, p (γ )/γ p (γ )/γ dt 4 dt k(t T m) dt k(t 75) dt dt T 75 kdt ln T 75 kt + c T (t) 75 + ce kt T () 5 c 6 so T 75 + 6e kt T () 95 95 75 + 6e k k ln T (t) 75 + 6e t ln Now if T (t) 65 then 65 75 + 6 t ln t h Thus the object was placed in the room at pm 5 dt dt k(t 45) T (t) 45 + Ce kt T () 5 C 4 so T (t) 45 4e kt and T () 5 k ln 4 ; hence, T (t) 45 4( 4 )t/ (i) T (4) 45 4( 4 ) 5 F (ii) T (t) 5 45 4( 4 )t/ ( 4 )t/ 4 ln 4 t 964 minutes ln(4/) dt 6 k(t 4) dt dt T 4 kdt T (t) 4 + ce kt T () 8 c 4 so that T (t) 4 + 4e kt T () 6 k ln ; hence, T (t) 4 + 4e t ln Now T (t) 98 T (t) 4 + 4e kt 98 t 6 t 4h Thus T ( 4) 98 and Holmes was right, the time of death was am 7 T (t) 75 + ce kt T () 45 75 + ce k 45 4 ce k and T () 47 75 + ce k 47 7 ce k Solving these two equations yields k ln 5 4 and c 45; hence, T 75 + 45( 4 5 )t/ (a) Furnace temperature: T () 5 F (b) If T (t) then 75 + 45( 4 ln 7 5 )t/ t ln 5 696 minutes Thus the temperature of 4 the coal was F at 6:7 pm 8 dt k(t 7) dt dt T 7 kdt T (t) 7 + ce kt Since dt, k(t 7) or dt k 9 Since T () 5 5 7 + ce /9 c 78e /9 ; consequently, T (t) 7 + 78e ( t)/9 (i) Initial temperature of the object: t T (t) 7 + 78e / 7 F (ii)rate of change of the temperature after minutes: T () 7 + 78e / so after minutes, dt dt 9 (7 + 78e / 7) dt dt 6 e / F per minute Solutions to Section 5

8 True-False Review: TRUE The differential equation for such a population growth is dp dt kp, where P (t) is the population as a function of time, and this is the Malthusian growth model described at the beginning of this section FALSE The initial population could be greater than the carrying capacity, although in this case the population will asymptotically decrease towards the value of the carrying capacity TRUE The differential equation governing the logistic model is (5), which is certainly separable as D dp P (C P ) dt r Likewise, the differential equation governing the Malthusian growth model is dp dt kp, and this is separable as dp P dt k 4 TRUE As (5) shows, as t, the population does indeed tend to the carrying capacity C independently of the initial population P As it does so, its rate of change dp dt slows to zero (this is best seen from (5) with P C) 5 TRUE Every five minutes, the population doubles (increase -fold) Over minutes, this population will double a total of 6 times, for an overall 6 64-fold increase 6 TRUE An 8-fold increase would take years, and a 6-fold increase would take 4 years Therefore, a -fold increase would take between and 4 years 7 FALSE The growth rate is dp dt constant kp, and so as P changes, dp dt changes Therefore, it is not always 8 TRUE From (5), the equilibrium solutions are P (t) and P (t) C, where C is the carrying capacity of the population 9 FALSE If the initial population is in the interval ( C, C), then although it is less than the carrying capacity, its concavity does not change To get a true statement, it should be stated instead that the initial population is less than half of the carrying capacity TRUE Since P (t) kp, then P (t) kp (t) k P > for all t Therefore, the concavity is always positive, and does not change, regardless of the initial population Problems: dp dt kp P (t) P e kt Since P (), then P e kt Since P (), then e k k ln Thus P (t) e(t/) ln Therefore, P (4) e (4/) ln 8 56 bacteria Using P (t) P e kt we obtain P () 5 5 p e k and P () 6 6 P e k which implies that e k 6 5 k ln 6 5 Hence, P () 5( 5 6 )5 94 Also, P P when t ln ln ln 6 76h 5 From P (t) P e kt and P () it follows that P (t) e kt Since t d 4, k 4 ln so P e t ln /4 Therefore, P (t) 6 6 e t ln /4 t 586h 4 dp dt kp P (t) P e kt Since, P () then P (t) e kt Since P (5) then

9 e kt k 5 ln Hence P (t) e(t ln )/5 (a) P () e 4 ln 6 (b) e (t ln )/5 e (t ln )/5 5 ln t yrs ln 5C 5 P (t) 5 + (C 5)e rt In formulas (55) and (56) we have P 5, P 8, P, t 5, and t Hence, r ()() 5 ln 8(8)(5) (5)() ln, C (5)() 5 8 486, so (5)() 486)(5) that P (t) 5 + 6486e t ln 574 Inserting t 5 into the preceding formula 5 + 6486e t ln yields P (5) 9 5C 6 P (t) 5 + (C 5)e rt In formulas (55) and (56) we have P 5, P 6, P 76, t, and t t 4 Hence, r (76)() ln 6(6)(6) (5)(76), C (5)(4) 6 9877, so that (5)(76) 4965 P (t) Inserting t into the preceding formula yields P () 5 + 4877e t 7 From equation (55) r > requires P (P P ) P (P P ) > Rearranging the terms in this inequality and using the fact that P > P yields P > P P P + P Further, C > requires that P (P + P ) P P P P P > From P > P P P + P we see that the numerator in the preceding inequality is positive, and therefore the denominator must also be positive Hence in addition to P > P P P + P, we must also have P > P P 8 Let y(t) denote the number of passengers who have the flu at time t Then we must solve dy dt ky(5 y), y() 5, y(), where k is a positive constant Separating the differential equation and integrating yields y(5 y) dy k dt Using a partial fraction decomposition on the left-hand side gives ( ) 5y + dy kt + c, so that 5(5 y) 5 ln y kt + c, which upon 5 y y exponentiation yields 5 y c e 5kt Imposing the initial condition y() 5, we find that c 99 y Hence, 5 y 99 e5kt The further condition y() requires for k gives k 99 ln 5 49 Therefore, y 5 y y(t) 5et ln (99/49) 99 + e t ln (99/49) 5 9 (a) Equilibrium solutions: P (t), P (t) T Slope: P > T dp dp >, < P < T dt dt < 49 99 e5k solving 99 et ln (99/49) Solving algebraically for y we find + 99e Hence, y(4) 5 474 t ln (99/49) + 99e 4 ln (99/49)

4 ( Isoclines: r(p T ) k P T P k r P ) rt + 4k T ± We see that slope of the r solution curves satisfies k rt 4 d P Concavity: dt r(p T )dp dt r (P T )(P T )P Hence, the solution curves are concave up for P > t, and are concave down for T, there is a population growth The term threshold level is appropriate since T gives the minimum value of P above which there is a population growth (a) Separating the variables in differential equation (57) gives in the equivalent form T (P T ) dp T P dt c e T rt The initial condition P () P requires P T P T P dp P (P T ) dt r Integrating yields T ln ( P T P P c, so that P T P r, which can be written ) rt + c, so that ( ) P T e rt t T P Solving algebraically for P yields P (t) P (P T )e rt t T P (b) If P < T, then the denominator in P (P T )e rt t is positive, and increases without bound as t Consequently lim t P (t) In this case the population dies out as t increases T P (c) If P > T, then the denominator of P (P T )e rt t vanishes when (P T )e rt t P, that is when ( P ) This means that within a finite time the population grows without bound We can t rt ln P T interpret this as a mathematical model of a population explosion dp dt r(c P )(P T )P, P () P, r >, < T < C Equilibrium solutions: P (t), P (t) T, P (t) C The slope of the solution curves is negative for C It is positive for T < P < C P

d P Concavity: dt r (C P )(P T ) (P T )P + (C P )P (C P )(P T )P, which simplifies to d P dt r ( P + P T + CP CT )(C P )(P T ) Hence changes in concavity occur when P (C + T ± C CT + T ) A representative slope field with some solution curves is shown in the acompanying figure We see that for C, then the solution decays towards the equilibrium solution P (t) C 4 P(t) C T t Figure 4: Figure for Exercise dp dt rp (ln C ln P ), P () P, and r, C, and P are positive constants Equilibrium solutions: P (t) C The slope of the solution curves is positive for C Concavity: d P dt r ln ( ) C dp P dt r ln ( ) C P ln C Hence, the solution curves are concave P P up for C They are concave down for C < P < C A representative slope field with e some solution curves are shown in the acompanying figure 5 5 P(t) 5 4 5 t Figure 44: Figure for Exercise

4 dp Separating the variables in (58) yields r which can be integrated directly to P (ln C ln P ) dt obtain ln (ln C ln P ) rt + c so that ln ( C P ) c e rt The initial condition P () P requires that ln ( C P ) c Hence, ln ( C P ) e rt ln ( C P ) so that P (t) Ce ln (P/k)e rt Since lim t e rt, it follows that lim t P (t) C 4 Using the exponential decay model we have dp dt kp, which is easily integrated to obtain P (t) P e kt The initial condition P () 4 requires that P 4, so that P (t) 4e kt We also know that P () 4 This requires that 4 4e k so that k ln ( 7 7 (a) From (), P (6)4e ln( ) 89 ln( 7 ) (b) From (), P () 4e (c) From (), the half-life, t H, is determine from 5 (a) More ) Consequently, P (t) 4e t 7 ln( ) () 4e t H ln( 7 ) th ln ln ( ) 8 days 7 (b) Using the exponential decay model we have dp dt kp, which is easily integrated to obtain P (t) P e kt The initial condition P (), requires that P,, so that P (t), e kt We also know that P () 8, This requires that, 8, e k so that k ln Using (4), the half-life is determined from ( 4 5 ) Consequently, P (t), e t ln( 4 5) (4) 5,, e t H ln( 4 5) th ln ln ( ) 5 6 min 4 (c) Using (4) there will be 5, fans left in the stadium at time t, where 5,, e t ln( 4 5) t ln ( ) ln ( ) 4 85 min 5 6 Using the exponential decay model we have dp dt kp, which is easily integrated to obtain P (t) P e kt Since the half-life is 5 years, we have Therefore, P P e 5k k ln 5 ln t P (t) P e 5

4 Consequently, only 4% of the original amount will remain at time t where 4 P P e t ln 5 t 5 ln 5 ln 45 years 7 Maple, or even a TI 9 plus, has no problem in solving these equations 8 (a) Malthusian model is P (t) 5e kt Since P () 794, then 794 5e k k t ln (794/5) Hence, P (t) 5e 5C (b)p (t) 5 + (C 5)e ln 794 5 rt Imposing the initial conditions P () 794 and P () ln (794/5) ln (794/5) gives the pair of equations 794 5e and 5e whose solution is 9956 C 695, r 46 Using these values for C and r gives P (t) 5 + 65e 46t (c) Malthusian model: P () 5 million; P (4) million P(t) 8 6 4 8 6 4 t Figure 45: Figure for Exercise 8(c) Logistics model: P () million; P (4) 6 million The logistics model fits the data better than the Malthusian model, but still gives a significant underestimate of the 99 population 5C 9 P (t) Imposing the conditions P (5), P (5) 5 gives the pair of equations 5 + (C 5)e rt 5C 5 + (C 5)e 5r and 5 5C whose positive solutions are C 7, r 7 5 + (C 5)e 5r 85 Using these values for C and r gives P (t) From the figure we see that it will take 5 + 845e 7t approximately 5 years to reach 95% of the carrying capacity Solutions to Section 6 True-False Review: FALSE Any solution to the differential equation (67) serves as an integrating factor for the differential

44 P(t) 5 Carrying Capacity 5 5 5 4 5 6 7 t Figure 46: Figure for Exercise 9 equation There are infinitely many solutions to (67), taking the form I(x) c e p(x)dx, where c is an arbitrary constant TRUE Any solution to the differential equation (67) serves as an integrating factor for the differential equation There are infinitely many solutions to (67), taking the form I(x) c e p(x)dx, where c is an arbitrary constant The most natural choice is c, giving the integrating factor I(x) e p(x)dx TRUE Multiplying y + p(x)y q(x) by I(x) yields y I + piy qi Assuming that I pi, the requirement on the integrating factor, we have y I + I y qi, or by the product rule, (I y) qi, as requested 4 FALSE Before determining an integrating factor, the equation must be rewritten as dy dx x y sin x, and with p(x) x x, we have integrating factor I(x) e dx, not e 5 TRUE Rewriting the differential equation as dy dx + x y x, we have p(x) x, and so an integrating factor must have the form I(x) e Since 5x does indeed have this form, it is an integrating factor x dx p(x)dx dx e x e ln x+c e c x Problems: In this section the function I(x) e p(x)dx will represent the integrating factor for a differential equation of the form y + p(x)y q(x) y y e x I(x) e dx e x d(e x y) dx e x e x y e x + c y(x) e x (e x + c) x y 4xy x 7 sin x, x > y 4 x y x5 sin x I(x) x 4 d(x 4 y) x sin x x 4 y dx sin x x cos x + c y(x) x 4 (sin x x cos x + c)

Solutions to Section 1.1