Unit 7 · 5–10% of the exam Beta

Unit 7 practice test: Differential Equations

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value or an expression; sets share one graph or table.

Covers: Writing Differential Equations; Verifying Solutions to Differential Equations; Sketching Slope Fields; Reasoning with Slope Fields; General Solutions by Separation of Variables; Particular Solutions from Initial Conditions; Exponential Models.

Graph

A solution of dydx=2y2\frac{dy}{dx} = 2y^2 through (0,12)\left(0, \frac{1}{2}\right)

−3−2−112−112345xy(0, 1/2)

1. Which is the particular solution of dydx=2y2\frac{dy}{dx} = 2y^2 with y(0)=12y(0) = \frac{1}{2}?

  1. y=12+2xy = \frac{1}{2 + 2x}
  2. y=−12x+12y = -\frac{1}{2x} + \frac{1}{2}
  3. y=21−xy = \frac{2}{1 - x}
  4. y=12−2xy = \frac{1}{2 - 2x}
Show the answer

y−2dy=2dxy^{-2}\,dy = 2\,dx, so −1y=2x+C-\frac{1}{y} = 2x + C. At (0,12)(0, \frac{1}{2}): −2=C-2 = C. Then 1y=2−2x\frac{1}{y} = 2 - 2x.

  • y=12+2xy = \frac{1}{2 + 2x}: ∫y−2dy=−1y\int y^{-2}\,dy = -\frac{1}{y}: the minus sign flips the 2x2x.
  • y=−12x+12y = -\frac{1}{2x} + \frac{1}{2}: The constant was added after solving for yy.
  • y=21−xy = \frac{2}{1 - x}: At x=0x = 0 this is 2, not 12\frac{1}{2}.
  • Correct: y=12−2xy = \frac{1}{2 - 2x}: −1y=2x+C-\frac{1}{y} = 2x + C; at (0,12)(0, \frac{1}{2}), C=−2C = -2, so y=12−2xy = \frac{1}{2 - 2x}.

2. What is the domain of this particular solution?

  1. x≠1x \ne 1
  2. x>1x > 1
  3. x<1x < 1
  4. (−∞,∞)(-\infty, \infty)
Show the answer

y=12−2xy = \frac{1}{2 - 2x} is undefined at x=1x = 1. The largest open interval containing x=0x = 0 is x<1x < 1.

  • x≠1x \ne 1: A solution's domain is one interval; the piece with x>1x > 1 is not connected to (0,12)(0, \frac{1}{2}).
  • x>1x > 1: The initial point x=0x = 0 must be in the domain.
  • Correct: x<1x < 1: The formula breaks at x=1x = 1; the solution lives on the interval containing x=0x = 0.
  • (−∞,∞)(-\infty, \infty): At x=1x = 1 the formula divides by 0.

3. What happens to this solution as x→1−x \to 1^-?

  1. yy increases without bound.
  2. yy approaches 12\frac{1}{2}.
  3. yy approaches 0.
  4. yy decreases without bound.
Show the answer

As x→1−x \to 1^-, 2−2x→02 - 2x \to 0 through positive values, so y→∞y \to \infty: a vertical asymptote ends the solution.

  • Correct: yy increases without bound.: 2−2x→0+2 - 2x \to 0^+, so 12−2x→∞\frac{1}{2 - 2x} \to \infty.
  • yy approaches 12\frac{1}{2}.: 12\frac{1}{2} is the value at x=0x = 0, not near x=1x = 1.
  • yy approaches 0.: The denominator shrinks to 0, so the fraction grows.
  • yy decreases without bound.: For x<1x < 1 the denominator is positive.

Graph

The slope field of dydx=x(2−y)\frac{dy}{dx} = x(2 - y)

−2−112−11234xy

4. A solution curve passes through (1,0)(1, 0). Near that point, the solution is

  1. decreasing
  2. constant
  3. at a relative maximum
  4. increasing
Show the answer

dydx=x(2−y)=2\frac{dy}{dx} = x(2 - y) = 2 at (1,0)(1, 0), so the solution is increasing there.

  • decreasing: At (1,0)(1, 0) both factors are positive.
  • constant: The slope is 2, not 0; constant solutions sit on y=2y = 2.
  • at a relative maximum: A relative maximum needs dydx=0\frac{dy}{dx} = 0 there.
  • Correct: increasing: dydx=1(2−0)=2>0\frac{dy}{dx} = 1(2 - 0) = 2 > 0.

5. The solution curve through (0,0)(0, 0) has dydx=0\frac{dy}{dx} = 0 at x=0x = 0. What happens there?

  1. A relative maximum: dydx\frac{dy}{dx} changes from positive to negative at x=0x = 0.
  2. Neither: dydx=0\frac{dy}{dx} = 0 but does not change sign.
  3. A relative minimum: dydx\frac{dy}{dx} changes from negative to positive at x=0x = 0.
  4. An equilibrium solution, since the slope is 0
Show the answer

Near (0,0)(0, 0), y<2y < 2, so dydx=x(2−y)\frac{dy}{dx} = x(2 - y) is negative for x<0x < 0 and positive for x>0x > 0: a relative minimum.

  • A relative maximum: dydx\frac{dy}{dx} changes from positive to negative at x=0x = 0.: Left of the yy-axis x<0x < 0 makes the slope negative, not positive.
  • Neither: dydx=0\frac{dy}{dx} = 0 but does not change sign.: It does change sign, with xx.
  • Correct: A relative minimum: dydx\frac{dy}{dx} changes from negative to positive at x=0x = 0.: For y<2y < 2, 2−y>02 - y > 0, so dydx\frac{dy}{dx} has the sign of xx.
  • An equilibrium solution, since the slope is 0: The slope is 0 at one point of the curve, not along a whole horizontal line.

6. For the solution curve through (0,0)(0, 0), what happens to yy as xx increases without bound?

  1. yy increases without bound.
  2. yy approaches 2.
  3. yy approaches 0.
  4. yy decreases without bound.
Show the answer

Below y=2y = 2 with x>0x > 0, dydx>0\frac{dy}{dx} > 0, and y=2y = 2 is an equilibrium solution the curve cannot cross, so y→2y \to 2.

  • yy increases without bound.: The curve cannot cross the equilibrium solution y=2y = 2.
  • Correct: yy approaches 2.: For x>0x > 0 and y<2y < 2 the curve rises, and the slopes flatten near the equilibrium y=2y = 2, which it cannot cross.
  • yy approaches 0.: For x>0x > 0, dydx>0\frac{dy}{dx} > 0 below y=2y = 2: the curve rises away from 0.
  • yy decreases without bound.: For x>0x > 0 and y<2y < 2 the slopes are positive.

7. A population of fish F decreases at a rate proportional to the square root of the population. With k > 0, which equation models this?

  1. dFdt=kF\frac{dF}{dt} = k\sqrt{F}
  2. dFdt=−kt\frac{dF}{dt} = -k\sqrt{t}
  3. dFdt=−kF\frac{dF}{dt} = -k\sqrt{F}
  4. F=−ktF = -k\sqrt{t}
Show the answer

"Decreases at a rate proportional to F\sqrt{F}": dFdt=−kF\frac{dF}{dt} = -k\sqrt{F} with k>0k > 0.

  • dFdt=kF\frac{dF}{dt} = k\sqrt{F}: With k>0k > 0 this rate is positive: the population would grow.
  • dFdt=−kt\frac{dF}{dt} = -k\sqrt{t}: The rate depends on the population, not on the time.
  • Correct: dFdt=−kF\frac{dF}{dt} = -k\sqrt{F}: Proportional to F\sqrt{F}, and the minus sign with k>0k > 0 makes the population decrease.
  • F=−ktF = -k\sqrt{t}: The sentence is about the rate of change of F, so a derivative is needed.

8. Which differential equation does y=sin⁡(2x)y = \sin(2x) satisfy?

  1. d2ydx2=4y\frac{d^2y}{dx^2} = 4y
  2. d2ydx2=−2y\frac{d^2y}{dx^2} = -2y
  3. d2ydx2=−4y\frac{d^2y}{dx^2} = -4y
  4. dydx=2y\frac{dy}{dx} = 2y
Show the answer

y′=2cos⁡(2x)y' = 2\cos(2x), y″=−4sin⁡(2x)=−4yy'' = -4\sin(2x) = -4y.

  • d2ydx2=4y\frac{d^2y}{dx^2} = 4y: The second derivative of sine brings a minus sign.
  • d2ydx2=−2y\frac{d^2y}{dx^2} = -2y: The chain rule gives a factor 2 twice: 2⋅2=42 \cdot 2 = 4.
  • Correct: d2ydx2=−4y\frac{d^2y}{dx^2} = -4y: y″=−4sin⁡(2x)=−4yy'' = -4\sin(2x) = -4y.
  • dydx=2y\frac{dy}{dx} = 2y: dydx=2cos⁡(2x)\frac{dy}{dx} = 2\cos(2x), not 2sin⁡(2x)2\sin(2x).

9. Where are the segments of the slope field for dydx=y2−1\frac{dy}{dx} = y^2 - 1 horizontal?

  1. On the line y=1y = 1 alone
  2. On the lines x=1x = 1 and x=−1x = -1
  3. On the lines y=1y = 1 and y=−1y = -1
  4. On the xx-axis
Show the answer

Horizontal segments have slope 0: y2−1=0y^2 - 1 = 0 gives y=1y = 1 or y=−1y = -1, for every xx.

  • On the line y=1y = 1 alone: (−1)2−1=0(-1)^2 - 1 = 0 as well.
  • On the lines x=1x = 1 and x=−1x = -1: The slope depends on yy, not xx.
  • Correct: On the lines y=1y = 1 and y=−1y = -1: y2−1=0y^2 - 1 = 0 when y=±1y = \pm 1.
  • On the xx-axis: At y=0y = 0 the slope is −1-1.

10. Separate and integrate dTdt=−0.05(T−20)\frac{dT}{dt} = -0.05(T - 20).

  1. ln⁡|T|−20=−0.05t+C\ln|T| - 20 = -0.05t + C
  2. ln⁡|T−20|=−0.05t\ln|T - 20| = -0.05t
  3. T−20=−0.05t+CT - 20 = -0.05t + C
  4. ln⁡|T−20|=−0.05t+C\ln|T - 20| = -0.05t + C
Show the answer

dTT−20=−0.05dt\frac{dT}{T - 20} = -0.05\,dt, so ln⁡|T−20|=−0.05t+C\ln|T - 20| = -0.05t + C.

  • ln⁡|T|−20=−0.05t+C\ln|T| - 20 = -0.05t + C: The whole T−20T - 20 is in the denominator: ∫dTT−20\int \frac{dT}{T - 20}=ln⁡|T−20|= \ln|T - 20|.
  • ln⁡|T−20|=−0.05t\ln|T - 20| = -0.05t: The constant of integration is missing.
  • T−20=−0.05t+CT - 20 = -0.05t + C: An antiderivative of 1T−20\frac{1}{T - 20} is a logarithm.
  • Correct: ln⁡|T−20|=−0.05t+C\ln|T - 20| = -0.05t + C: ∫dTT−20\int \frac{dT}{T - 20}=∫−0.05dt= \int -0.05\,dt.

11. A quantity satisfies dydt=0.08y\frac{dy}{dt} = 0.08y and y(0)=50y(0) = 50. Which is y(t)y(t)?

  1. y=0.08e50ty = 0.08e^{50t}
  2. y=50+e0.08ty = 50 + e^{0.08t}
  3. y=50e0.08ty = 50e^{0.08t}
  4. y=50e−0.08ty = 50e^{-0.08t}
Show the answer

dydt=ky\frac{dy}{dt} = ky with y(0)=y0y(0) = y_0 gives y=y0ekt=50e0.08ty = y_0e^{kt} = 50e^{0.08t}.

  • y=0.08e50ty = 0.08e^{50t}: The starting value and the constant are swapped.
  • y=50+e0.08ty = 50 + e^{0.08t}: At t=0t = 0 this is 51, and its rate is not 0.08y0.08y.
  • Correct: y=50e0.08ty = 50e^{0.08t}: y=y0ekty = y_0e^{kt} with y0=50y_0 = 50, k=0.08k = 0.08.
  • y=50e−0.08ty = 50e^{-0.08t}: k=+0.08k = +0.08: growth, not decay.

12. A quantity satisfies dydt=0.4(50−y)\frac{dy}{dt} = 0.4(50 - y). For which values of yy is yy increasing?

  1. y>50y > 50
  2. t<50t < 50
  3. y>0y > 0
  4. y<50y < 50
Show the answer

yy increases where dydt>0\frac{dy}{dt} > 0: 0.4(50−y)>00.4(50 - y) > 0 exactly when y<50y < 50.

  • y>50y > 50: Then 50−y<050 - y < 0 and dydt<0\frac{dy}{dt} < 0: yy decreases.
  • t<50t < 50: The sign of the rate depends on y, not on t.
  • y>0y > 0: At y=60y = 60, which is positive, dydt=−4<0\frac{dy}{dt} = -4 < 0.
  • Correct: y<50y < 50: Then 50−y>050 - y > 0, so dydt>0\frac{dy}{dt} > 0.

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Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.