Covers Lessons 11–15: differential equations, separable equations, growth and decay models, motion applications, and module synthesis.
Assessment
Select the best answer for each question.
The order of $\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 0$ is:
$\frac{dy}{dx} = xy$ is:
The solution to $\frac{dN}{dt} = kN$ with $N(0) = N_0$ is:
A substance with half-life 10 years has decay constant:
If $v(t) = \frac{dx}{dt}$, then $x(t)$ equals:
$\frac{dy}{dx} = x + y$ is:
The doubling time for $\frac{dP}{dt} = 0.05P$ is approximately:
Total distance travelled by a particle is found by:
Short Answer
Solve $\frac{dy}{dx} = 2xy$ with $y(0) = 3$. Show all working.
A particle has acceleration $a(t) = 3t^2$ m/s². At $t = 0$, $v = 2$ m/s and $x = 5$ m. Find $x(2)$. Show all working.
A bacterial colony grows from 500 to 2000 in 8 hours. (a) Find the growth constant $k$. (b) How many bacteria after 12 hours? (c) When does the population reach 10,000? Show all working.
Q1: C — Second order (highest derivative is $\frac{d^2y}{dx^2}$).
Q2: A — $\frac{dy}{dx} = xy$ can be written as $\frac{dy}{y} = x\,dx$.
Q3: B — $N = N_0 e^{kt}$.
Q4: C — $t_{1/2} = \frac{\ln 2}{|k|}$, so $|k| = \frac{\ln 2}{10}$.
Q5: B — Position is the integral of velocity.
Q6: B — $x + y$ cannot be factored as $f(x)g(y)$.
Q7: C — $t_{double} = \frac{\ln 2}{0.05} \approx 13.9$ years.
Q8: B — Distance uses $|v(t)|$ to account for backward motion.
Q9 (3 marks): $\frac{dy}{y} = 2x\,dx$ [0.5]. $\ln|y| = x^2 + C$ [1]. $y = Ae^{x^2}$ [0.5]. $y(0) = 3$: $A = 3$ [0.5]. $y = 3e^{x^2}$ [0.5].
Q10 (3 marks): $v(t) = t^3 + C_1$ [0.5]. $v(0) = 2$: $C_1 = 2$ [0.5]. $x(t) = \frac{t^4}{4} + 2t + C_2$ [0.5]. $x(0) = 5$: $C_2 = 5$ [0.5]. $x(2) = 4 + 4 + 5 = 13$ m [1].
Q11 (4 marks): (a) $2000 = 500e^{8k}$ [0.5], $e^{8k} = 4$, $k = \frac{\ln 4}{8} = \frac{\ln 2}{4} \approx 0.173$ per hour [1]. (b) $P(12) = 500e^{0.173 \times 12} = 500e^{2.079} \approx 4000$ bacteria [1.5]. (c) $10000 = 500e^{kt}$, $e^{kt} = 20$, $t = \frac{\ln 20}{k} \approx 17.3$ hours [1].