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Solve the problem. -The system of equations dxdt=(2+3y) x\frac { d x } { d t } = ( - 2 + 3 y ) x and dydt=(2+x) y\frac { d y } { d t } = ( - 2 + x ) y describes the growth rates of two symbiotic (dependent) species of animals (such as the rhinoceros and a type of bird which eats insects from its back) . What is necessarily true of the two populations at the equilibrium points?


A) They are both at a maximum.
B) Both populations equal zero.
C) They are both at a minimum.
D) They both remain constant over all time.

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Obtain a slope field and add to its graphs of the solution curves passing through the given points. - y=3(y1) with (2,0)y ^ { \prime } = 3 ( y - 1 ) \text { with } ( 2,0 )

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Solve the problem. -A 200 gal tank is half full of distilled water. At time = 0, a solution containing 1 lb/gal of concentrate enters the tank at the rate of 4 gal/min, and the well-stirred mixture is withdrawn at the rate of 2 gal/min. When the tank Is full, how many pounds of concentrate will it contain?


A) 100 pounds
B) 200 pounds
C) 120 pounds
D) 150 pounds

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The autonomous differential equation represents a model for population growth. Use phase line analysis to sketch solution curves for P(t), selecting different starting values P(0). Which equilibria are stable, and which are unstable? - dPdt=P(P5)\frac { \mathrm { dP } } { \mathrm { dt } } = \mathrm { P } ( \mathrm { P } - 5 )

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Determine which of the following equations is correct. - 1sinxsinxdx=\frac { 1 } { \sin x } \int \sin x d x =


A) 1/sinx+C1 / \sin x + C
B) cotx+C- \cot x + C
C) cosx+C\cos x + C
D) sinx+C\sin x + C

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Solve the differential equation. - 3x2y2xy=y33 x ^ { 2 } y ^ { \prime } - 2 x y = y ^ { - 3 }


A) y4=411x+Cx8/3y ^ { 4 } = \frac { - 4 } { 11 x } + C x ^ { 8 / 3 }
B) y4=411x+Cx11/3y ^ { 4 } = \frac { - 4 } { 11 x } + C x ^ { 11 / 3 }
C) y2=411x+Cx11/3y ^ { 2 } = \frac { - 4 } { 11 x } + C x ^ { 11 / 3 }
D) y2=411x+Cx8/3y ^ { 2 } = \frac { - 4 } { 11 x } + C x ^ { 8 / 3 }

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Solve the differential equation. - exdydx+3exy=2,x>0e ^ { x } \frac { d y } { d x } + 3 e ^ { x } y = 2 , x > 0


A) y=e3x+Cex,x>0y = e ^ { - 3 x } + C e ^ { - x } , x > 0
B) y=ex+e3x,x>0y = e ^ { - x } + e ^ { - 3 x } , x > 0
C) y=ex+Ce3x,x>0y = e ^ { - x } + C e ^ { - 3 x } , x > 0
D) y=ex+Ce3x,x>0y = e ^ { x } + C e ^ { - 3 x } , x > 0

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Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Round your results to four decimal places. - y=4xex4,y(1) =3,dx=0.1y ^ { \prime } = 4 x e ^ { x ^ { 4 } } , y ( 1 ) = 3 , d x = 0.1


A) y1 = 3.2699, y2 = 5.1723, y3 = 8.9899
B) y1 = 4.0873, y2 = 5.9897, y3 = 9.8074
C) y1 = 3.6786, y2 = 5.5810, y3 = 9.3986
D) y1 = 4.4960, y2 = 6.3985, y3 = 10.2161

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B

Solve the initial value problem. - dydx+xy=4x;y(0) =4\frac { d y } { d x } + x y = 4 x ; y ( 0 ) = - 4


A) y=4ex2/28y = 4 e ^ { x ^ { 2 } / 2 } - 8
B) y=4ex2/28y = 4 e ^ { - x ^ { 2 } / 2 } - 8
C) y=8ex2/2+4y = - 8 e ^ { x ^ { 2 } / 2 } + 4
D) y=8ex2/2+4y = - 8 e ^ { - x ^ { 2 } / 2 } + 4

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D

Sketch several solution curves. - dydx=y21\frac { d y } { d x } = y ^ { 2 } - 1

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Identify equilibrium values and determine which are stable and which are unstable. - dydx=y216\frac { \mathrm { dy } } { \mathrm { dx } } = \mathrm { y } ^ { 2 } - 16


A) y = 4 is a stable equilibrium value and y = -4 is an unstable equilibrium.
B) y = -4 and y = 5 are stable equilibrium values.
C) y = -4 is a stable equilibrium value and y = 4 is an unstable equilibrium.
D) There are no equilibrium values.

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Identify equilibrium values and determine which are stable and which are unstable. -y = (y - 4) (y - 6) (y - 7)


A) y = 7 is a stable equilibrium value and y = 6 and y = 4 are unstable equilibria.
B) y = 4 is a stable equilibrium value and y = 6 and y = 7 are unstable equilibria.
C) y = 6 is a stable equilibrium value and y = 4 and y = 7 are unstable equilibria.
D) y = 5, y = 4 and y = 7 are unstable equilibria.

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Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = 11t.


A) y=550t+27,500+37,500e.02t\mathrm { y } = 550 \mathrm { t } + 27,500 + 37,500 \mathrm { e } ^ { - .02 \mathrm { t } }
B) y=550t27,500+37,500e.02ty = 550 \mathrm { t } - 27,500 + 37,500 \mathrm { e } ^ { - .02 \mathrm { t } }
C) y=550t27,500+37,500e.02ty = - 550 t - 27,500 + 37,500 e ^ { - .02 t }
D) y=550t27,500+37,500e02ty = - 550 \mathrm { t } - 27,500 + 37,500 \mathrm { e }^{ \cdot 02 \mathrm { t }}

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Solve the problem. -dy/dt = ky + f(t) is a population model where y is the population at time t and f(t) is some function to describe the net effect on the population. Assume k = .02 and y = 10,000 when t = 0. Solve the differential equation of y When f(t) = -8t.


A) y=400t20,00010,000e02ty = - 400 t - 20,000 - 10,000 \mathrm { e } \cdot 02 \mathrm { t }
B) y=400t+20,00010,000e.02ty = - 400 t + 20,000 - 10,000 \mathrm { e } ^ { - .02 t }
C) y=400t+20,00010,000e.02ty = 400 t + 20,000 - 10,000 e ^ { - .02 t }
D) y=400t+20,00010,000e02ty = 400 t + 20,000 - 10,000 \mathrm { e } ^{\cdot 02 \mathrm { t }}

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Find the orthogonal trajectories of the family of curves. Sketch several members of each family. -y = -mx

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Construct a phase line. Identify signs of y and y . - dydx=(y+1) (y2) \frac{d y}{d x}=(y+1) (y-2)


A)
 Construct a phase line. Identify signs of y and y . - \frac{d y}{d x}=(y+1) (y-2)   A)     B)     C)     D)

B)
 Construct a phase line. Identify signs of y and y . - \frac{d y}{d x}=(y+1) (y-2)   A)     B)     C)     D)

C)
 Construct a phase line. Identify signs of y and y . - \frac{d y}{d x}=(y+1) (y-2)   A)     B)     C)     D)

D)
 Construct a phase line. Identify signs of y and y . - \frac{d y}{d x}=(y+1) (y-2)   A)     B)     C)     D)

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Obtain a slope field and add to its graphs of the solution curves passing through the given points. - y=2yx with (2,0)\mathrm { y } ^ { \prime } = \frac { 2 \mathrm { y } } { \mathrm { x } } \text { with } ( - 2,0 )

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Solve. -Solve the initial value problem dPdt=7kP2,P(0) =P0\frac { \mathrm { dP } } { \mathrm { dt } } = 7 \mathrm { kP } ^ { 2 } , \mathrm { P } ( 0 ) = \mathrm { P } _ { 0 }


A) P(t) =7P017kP0tP ( t ) = \frac { 7 P _ { 0 } } { 1 - 7 \mathrm { kP } _ { 0 } t }
B) P(t) =P017kP0t\mathrm { P } ( \mathrm { t } ) = \frac { \mathrm { P } _ { 0 } } { 1 - 7 \mathrm { kP } _ { 0 } \mathrm { t } }
C) P(t) =P077kP0t\mathrm { P } ( \mathrm { t } ) = \frac { \mathrm { P } _ { 0 } } { 7 - 7 \mathrm { kP } _ { 0 } \mathrm { t } }
D) P(t) =P01kP0t\mathrm { P } ( \mathrm { t } ) = \frac { \mathrm { P } _ { 0 } } { 1 - \mathrm { kP } _ { 0 } \mathrm { t } }

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Solve the problem. -An office contains 1000ft31000 \mathrm { ft } ^ { 3 } of air initially free of carbon monoxide. Starting at time =0= 0 , cigarette smoke containing 4%4 \% carbon monoxide is blown into the room at the rate of 0.5ft3/min0.5 \mathrm { ft } 3 / \mathrm { min } . A ceiling fan keeps the air in the room well circulated and the air leaves the room at the same rate of 0.5ft3/min. Find 0.5 \mathrm { ft } ^ { 3 } / \mathrm { min } ^ { \text {. Find } } the time when the concentration of carbon monoxide reaches 0.01%0.01 \% .


A) 5.01 min
B) 6.01 min
C) 8.01 min
D) 7.01 min

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A

Solve the problem. -How many seconds after the switch in an RL circuit is closed will it take the current i to reach 20% of its steady state value? Express answer in terms of R and L and round coefficient to the nearest hundredth.


A) 0.22 L/R seconds
B) 1.81 L/R seconds
C) 1.61 L/R seconds
D) 0.42 L/R seconds

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