E078 Section 5 — Practice Test (15 Q)

Section 5 (Differential Equations) practice test: 15 problems with final answers and condensed worked solutions. Companion to the E078 Section 5 flashcards. Not official WGU material.

Q1 (5.1). State the order of each differential equation. (a) $y'' + 3y' - 4y = e^x$ (b) $(y')^3 + y = x$ (c) $y^{(4)} - y'' = 0$

Answer: (a) order 2, (b) order 1, (c) order 4 | (a) Highest derivative is $y''$. | (b) Only $y'$ appears; it is cubed, but a power is not a derivative. | (c) $y^{(4)}$ is the highest derivative.

Q2 (5.1). Verify that $y = 4e^{-x} + x - 1$ is a solution of $y' + y = x$.

Answer: It is a solution (left side simplifies to $x$). | $y' = -4e^{-x} + 1$. | $y' + y = (-4e^{-x} + 1) + (4e^{-x} + x - 1) = x$, which equals the right side.

Q3 (5.1). Solve the initial-value problem $y' = 6x^2 - 4x + 1$, $y(1) = 3$.

Answer: $y = 2x^3 - 2x^2 + x + 2$ | Integrate: $y = 2x^3 - 2x^2 + x + C$. | $y(1) = 2 - 2 + 1 + C = 1 + C = 3$, so $C = 2$.

Q4 (5.2). For $y' = x^2 - 2y$, find the slope of the solution curve through each point: $(0,1)$, $(1,0)$, $(2,2)$.

Answer: $-2$, $1$, $0$ | Slope $= x^2 - 2y$. | At $(0,1)$: $0 - 2 = -2$. | At $(1,0)$: $1 - 0 = 1$. | At $(2,2)$: $4 - 4 = 0$ (a horizontal segment).

Q5 (5.1). A ball is thrown straight up from a height of 5 m with initial velocity 19.6 m/s; only gravity acts on it. (a) Find $v(t)$ and $s(t)$. (b) Find the maximum height. (c) When does it hit the ground, and how fast is it going?

Answer: (a) $v(t) = -9.8t + 19.6$, $s(t) = -4.9t^2 + 19.6t + 5$; (b) 24.6 m; (c) $t \approx 4.2406$ s, speed about 21.96 m/s (falling) | (a) $v' = -9.8$, $v(0) = 19.6$; then $s' = v$, $s(0) = 5$. | (b) Top when $v = 0$: $t = 2$ s, $s(2) = -19.6 + 39.2 + 5 = 24.6$. | (c) $s = 0$: $t = \frac{19.6 + \sqrt{19.6^2 + 4(4.9)(5)}}{9.8} \approx 4.2406$ (reject the negative root); $v(4.2406) \approx -21.958$ m/s.

Q6 (5.2). Find the equilibrium solutions of $y' = (y - 2)(y + 1)$ and classify each as stable, unstable, or semi-stable.

Answer: $y = -1$ is stable; $y = 2$ is unstable | Set $(y-2)(y+1) = 0$: $y = -1$, $y = 2$. | Sign chart: $y < -1$: $(-)(-) = +$ (up); $-1 < y < 2$: $(-)(+) = -$ (down); $y > 2$: $+$ (up). | $y = -1$: both sides move toward it (stable). $y = 2$: both sides move away (unstable).

Q7 (5.2). Use Euler's method with $h = 0.2$ to approximate $y(0.6)$ for $y' = 2x - y$, $y(0) = 1$. Show a table.

Answer: $y(0.6) \approx 0.736$ | $f(x,y) = 2x - y$, $y_{n+1} = y_n + 0.2 f(x_n, y_n)$. | $n=0$: $x=0$, $y=1$, $f=-1$, $y_1 = 0.8$. | $n=1$: $x=0.2$, $y=0.8$, $f=-0.4$, $y_2 = 0.72$. | $n=2$: $x=0.4$, $y=0.72$, $f=0.08$, $y_3 = 0.736$ at $x = 0.6$.

Q8 (5.3). Find the general solution of $\frac{dy}{dx} = y\cos x$.

Answer: $y = Ce^{\sin x}$ | Constant solution $y = 0$. | Separate: $\frac{dy}{y} = \cos x\,dx \Rightarrow \ln|y| = \sin x + C \Rightarrow |y| = e^C e^{\sin x}$. | Let the constant take any sign (and $0$ covers $y = 0$): $y = Ce^{\sin x}$.

Q9 (5.3). Solve $\frac{dy}{dx} = \frac{x}{y}$, $y(0) = -2$, as an explicit function.

Answer: $y = -\sqrt{x^2 + 4}$ | $y\,dy = x\,dx \Rightarrow \frac{y^2}{2} = \frac{x^2}{2} + C \Rightarrow y^2 = x^2 + C_1$. | $y(0) = -2$: $4 = C_1$, so $y = \pm\sqrt{x^2 + 4}$. | Since $y(0) = -2 < 0$, take the negative root.

Q10 (5.3). Solve $\frac{dy}{dt} = 3t^2(y + 1)$, $y(0) = 1$.

Answer: $y = 2e^{t^3} - 1$ | Constant solution $y = -1$ does not fit $y(0) = 1$. | Separate: $\frac{dy}{y+1} = 3t^2\,dt \Rightarrow \ln|y+1| = t^3 + C \Rightarrow y + 1 = C_1 e^{t^3}$. | $y(0) = 1$: $2 = C_1$.

Q11 (5.3). A 50 L tank starts with pure water. Brine containing 0.2 kg of salt per liter flows in at 5 L/min, and the well-mixed solution drains at 5 L/min. (a) Write the IVP for the salt $u(t)$ (kg). (b) Solve it. (c) Find the limiting amount. (d) When is there 5 kg of salt?

Answer: (a) $\frac{du}{dt} = 1 - \frac{u}{10}$, $u(0) = 0$; (b) $u(t) = 10 - 10e^{-t/10}$; (c) 10 kg; (d) $t = 10\ln 2 \approx 6.93$ min | (a) Rate in $= 0.2(5) = 1$ kg/min; rate out $= \frac{u}{50}(5) = \frac{u}{10}$. | (b) $\frac{du}{10 - u} = \frac{dt}{10} \Rightarrow -\ln|10 - u| = \frac{t}{10} + C \Rightarrow 10 - u = C_1 e^{-t/10}$; $u(0) = 0 \Rightarrow C_1 = 10$. | (c) $e^{-t/10} \to 0$, so $u \to 10$. | (d) $e^{-t/10} = \frac{1}{2}$.

Q12 (5.3). An object at $200^\circ\text{F}$ is placed in a room at $60^\circ\text{F}$. After 15 minutes it is $140^\circ\text{F}$. Find $T(t)$ and the time when it reaches $100^\circ\text{F}$.

Answer: $T(t) = 60 + 140e^{-0.037308t}$; reaches $100^\circ\text{F}$ at $t \approx 33.58$ min | $T_s = 60$, $T_0 = 200$: $T(t) = 60 + 140e^{kt}$. | $T(15) = 140$: $80 = 140e^{15k} \Rightarrow k = \frac{\ln(4/7)}{15} \approx -0.037308$. | $100 = 60 + 140e^{kt} \Rightarrow e^{kt} = \frac{2}{7} \Rightarrow t = \frac{\ln(2/7)}{k} = \frac{15\ln(2/7)}{\ln(4/7)} \approx 33.58$.

Q13 (5.4). A population has $P_0 = 50$, growth rate $r = 0.3$ per year, and carrying capacity $K = 500$. (a) Write the logistic IVP. (b) Solve it. (c) Find $P(10)$. (d) When is the population growing fastest, and what is $P$ then?

Answer: (a) $\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{500}\right)$, $P(0) = 50$; (b) $P(t) = \frac{500e^{0.3t}}{9 + e^{0.3t}}$; (c) $P(10) \approx 345.28$ (about 345); (d) $t = \frac{\ln 9}{0.3} \approx 7.32$ years, when $P = K/2 = 250$ | (b) $P(t) = \frac{P_0 K e^{rt}}{(K - P_0) + P_0 e^{rt}} = \frac{50(500)e^{0.3t}}{450 + 50e^{0.3t}}$. | (c) $P(10) = \frac{500e^3}{9 + e^3}$. | (d) Inflection at $t = \frac{1}{0.3}\ln\frac{450}{50}$.

Q14 (5.4). Consider $\frac{dP}{dt} = 0.1P\left(1 - \frac{P}{800}\right)$. (a) Find the equilibria and their stability. (b) For what $P$ is the population growing fastest, and what is that rate? (c) Describe what happens if $P(0) = 1000$.

Answer: (a) $P = 0$ unstable, $P = 800$ stable; (b) fastest at $P = 400$, max rate $20$ per unit time; (c) $P$ decreases toward 800 (never drops below 800) | (a) $0.1P(1 - P/800) = 0$: $P = 0$, $800$. Signs: $P < 0$: $-$; $0 < P < 800$: $+$; $P > 800$: $-$. | (b) Rate is a downward parabola in $P$, maximized halfway between the zeros at $P = 400$; max rate $= 0.1(400)(\frac{1}{2}) = 20 = rK/4$. | (c) $P > 800$ makes the right side negative.

Q15 (5.2 + 5.3). Consider $y' = y^2 - 1$. (a) Find the equilibria and classify their stability. (b) Solve the IVP with $y(0) = 0$. (c) What does $y$ approach as $x \to \infty$? Is that consistent with (a)?

Answer: (a) $y = -1$ stable, $y = 1$ unstable; (b) $y = \frac{1 - e^{2x}}{1 + e^{2x}}$; (c) $y \to -1$, yes, consistent | (a) $y^2 - 1 = 0$: $y = \pm 1$. Signs: $y < -1$: $+$; $-1 < y < 1$: $-$; $y > 1$: $+$. | (b) Partial fractions: $\frac{1}{2}\ln\left|\frac{y-1}{y+1}\right| = x + C \Rightarrow \frac{y-1}{y+1} = C_1 e^{2x}$; $y(0) = 0$ gives $C_1 = -1$; solve for $y$. | (c) Divide by $e^{2x}$: $y \to -1$, the stable equilibrium below the start.