This TSA checks first-order differential equations, separation of variables, substitution, finding unknowns from a given solution, formulating differential equations from word problems, interpreting solution behaviour, and solving a linked differential equation word problem.
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Core Ideas and First-Order Differential Equations
1. A general solution of a first-order differential equation usually contains an arbitrary constant, while a particular solution has the constant determined using given information.
Tested fundamental: Distinguishing general and particular solutions.
Explanation: A general solution represents a family of possible solutions. A particular solution is obtained when the arbitrary constant is fixed using a condition such as \(y=2\) when \(x=0\).
Common error: Treating a solution with an arbitrary constant as though it is already a particular solution.
2. Find the particular solution of \[ \sin^2 x\frac{dy}{dx}=\cos x, \] given that \[ y=0\quad\text{when}\quad x=\frac{\pi}{2}. \]
Tested fundamental: Solving a first-order differential equation by direct integration and applying an initial condition.
Explanation: First rewrite the DE as \(\frac{dy}{dx}=\frac{\cos x}{\sin^2 x}\). Then \(y=\int \frac{\cos x}{\sin^2 x}\,dx=-\mathrm{cosec}\,x+C\). Using \(y=0\) when \(x=\frac{\pi}{2}\), we get \(C=1\).
Common wrong choice: A has the sign of the integral reversed.
3. Which integral equation is obtained after separating variables for \[ \frac{dy}{dx}=x(1+y^2)? \]
Tested fundamental: Separating variables before integrating.
Explanation: Move all \(y\)-terms to the left and all \(x\)-terms to the right: \(\frac{1}{1+y^2}\,dy=x\,dx\). Therefore, \(\int \frac{1}{1+y^2}\,dy=\int x\,dx\).
Common wrong choice: B integrates with respect to \(x\) while still leaving \(y\)-terms inside the integrand.
4. Solve \[ \frac{dy}{dx}=xy, \] given that \(y=2\) when \(x=0\).
Tested fundamental: Solving a separable differential equation with an initial condition.
Explanation: Separating variables gives \(\frac1y\,dy=x\,dx\). So \(\ln|y|=\frac{x^2}{2}+C\), hence \(y=Ae^{x^2/2}\). Using \(y=2\) when \(x=0\), we get \(A=2\).
Common wrong choice: B comes from integrating \(x\) incorrectly as \(x\) instead of \(\frac{x^2}{2}\).
Substitution and Given Solutions
5. If \[ y=vx^2, \] where \(v\) and \(x\) are variables, which expression gives \(\frac{dy}{dx}\)?
Tested fundamental: Differentiating a substitution using the product rule.
Explanation: Using the product rule, \(\frac{dy}{dx}=x^2\frac{dv}{dx}+2xv\).
Common wrong choice: C treats \(v\) as a constant instead of a variable depending on \(x\).
6. Given \[ \frac{dy}{dx}=2x(1+y), \] use the substitution \[ u=x^2+1. \] Which differential equation in \(u\) and \(y\) is obtained?
Tested fundamental: Reducing a differential equation using a given substitution.
Explanation: Since \(u=x^2+1\), \(\frac{du}{dx}=2x\). Also, \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2x\frac{dy}{du}\). Comparing with \(\frac{dy}{dx}=2x(1+y)\), we get \(\frac{dy}{du}=1+y\).
Common wrong choice: A changes the derivative notation but does not properly apply the chain rule.
7. A differential equation is given by \[ \frac{dy}{dx}+y=ax+b, \] where \(a\) and \(b\) are constants. Its general solution is \[ y=3x-2+Ce^{-x}. \] Find \(a\) and \(b\).
Tested fundamental: Finding unknown constants in a differential equation from its given general solution.
Explanation: Differentiate \(y=3x-2+Ce^{-x}\) to get \(\frac{dy}{dx}=3-Ce^{-x}\). Then \(\frac{dy}{dx}+y=3-Ce^{-x}+3x-2+Ce^{-x}=3x+1\). So \(a=3,\ b=1\).
Common error: Forgetting that the \(Ce^{-x}\) terms cancel when the solution is substituted into the differential equation.
Formulating Differential Equations from Word Problems
8. A new social media app has a potential market of \(80\) thousand users. Let \(N\) be the number of users, in thousands, at time \(t\) months. The rate of increase of \(N\) is proportional to the product of the number of current users and the number of potential users who have not joined.
Which differential equation models this situation?
Tested fundamental: Forming a logistic-type differential equation from proportionality wording.
Explanation: The number of current users is \(N\), and the number of potential users who have not joined is \(80-N\). Since the rate is proportional to their product, \(\frac{dN}{dt}=kN(80-N)\).
Common wrong choice: A only accounts for potential users who have not joined, but ignores the current number of users.
9. A patient is given medicine through a drip at a constant rate \(r\). The medicine is removed from the bloodstream at a rate proportional to the amount \(M\) of medicine currently present.
Which differential equation models \(M\)?
Tested fundamental: Forming a net-rate differential equation from constant input and proportional output.
Explanation: The net rate is rate in minus rate out. The input rate is \(r\), and the removal rate is \(kM\). Hence \(\frac{dM}{dt}=r-kM\).
Common wrong choice: A reverses input and output, giving the wrong sign.
10. A 3D printer resin tray in the shape of a cuboid has a rectangular base area of \(40\text{ cm}^2\). Resin is pumped into the tray at a constant rate of \(24\text{ cm}^3\text{ min}^{-1}\). The printer uses up resin at a rate proportional to the current depth \(h\) cm of resin in the tray.
Which differential equation gives \(\frac{dh}{dt}\), where \(k>0\)?
Tested fundamental: Combining proportion, net rate and rate of change to formulate a differential equation.
Explanation: Let \(V\) be the volume of resin in the tray. Since the tray is cuboidal with rectangular base area \(40\text{ cm}^2\), \(V=40h\). So \(\frac{dV}{dt}=40\frac{dh}{dt}\). The net rate of change of volume is \(24-kh\). Hence \(40\frac{dh}{dt}=24-kh\), so \(\frac{dh}{dt}=\frac{24-kh}{40}\).
Common wrong choice: A gives the rate of change of volume, not the rate of change of height.
11. In a chemical reaction, \(2\) units of chemical \(A\) and \(1\) unit of chemical \(B\) are used to form \(1\) unit of product \(W\). Initially, there are \(40\) units of \(A\) and \(50\) units of \(B\). Let \(w\) be the amount of product formed at time \(t\). The rate of formation of \(W\) is proportional to the product of the amounts of \(A\) and \(B\) remaining.
Which differential equation models \(w\)?
Tested fundamental: Forming a differential equation from reaction amounts remaining.
Explanation: If \(w\) units of product have formed, then \(2w\) units of \(A\) and \(w\) units of \(B\) have been used. So the remaining amounts are \(40-2w\) and \(50-w\). Therefore, \(\frac{dw}{dt}=k(40-2w)(50-w)\).
Common wrong choice: B forgets that \(2\) units of \(A\) are used for each unit of product formed.
12. A bacteria population \(N\), in thousands, follows \[ \frac{dN}{dt}=kN(80-N). \] Initially, \(N=5\), and the initial growth rate is \(15\) thousand bacteria per hour. Find \(k\).
Tested fundamental: Finding a constant of proportionality from an initial rate.
Explanation: At \(t=0\), \(N=5\) and \(\frac{dN}{dt}=15\). Substitute into the DE: \(15=k(5)(80-5)=375k\). Hence \(k=\frac{1}{25}\).
Common error: Substituting only the initial value \(N=5\) but not the initial rate \(\frac{dN}{dt}=15\).
13. A parachutist falls with downward velocity \(v\). The downward acceleration due to gravity is \(9.8\), and the upward air resistance produces a deceleration proportional to \(v\), with constant \(k>0\).
Which statement is correct?
Tested fundamental: Forming and interpreting a motion differential equation with terminal velocity.
Explanation: Gravity increases downward velocity, while air resistance decreases it. Hence \(\frac{dv}{dt}=9.8-kv\). At terminal velocity, \(\frac{dv}{dt}=0\), so \(9.8-kv=0\), giving \(v=\frac{9.8}{k}\).
Common wrong choice: A has the wrong sign for air resistance, which would make the velocity grow without bound.
Interpreting Solutions and Long-Run Behaviour
14. The number of people who downloaded a game is modelled by \[ P=\frac{26}{9e^{-t/2}+4}, \] where \(P\) is measured in thousands. What is the long-run number of people who downloaded the game?
Tested fundamental: Interpreting long-run behaviour from a solution involving exponential decay.
Explanation: As \(t\to\infty\), \(e^{-t/2}\to0\). So \(P\to \frac{26}{4}=6.5\). Since \(P\) is measured in thousands, the long-run number is \(6500\) people.
Common wrong choice: D ignores that the limiting value is \(\frac{26}{4}\), not \(26\), and also mishandles the unit “thousands”.
15. A solution is given by \[ Q(t)=75(1-e^{-kt}), \] where \(k>0\). Which description best matches the graph of \(Q\) against \(t\) for \(t\ge0\)?
Tested fundamental: Interpreting the shape and horizontal asymptote of a solution curve.
Explanation: When \(t=0\), \(Q(0)=75(1-e^0)=0\). As \(t\to\infty\), \(e^{-kt}\to0\), so \(Q(t)\to75\). Thus the graph starts at \(0\), increases, and levels off at \(Q=75\).
Common wrong choice: C misses the horizontal asymptote and incorrectly assumes the quantity increases without bound.
Linked Scenario: Solving a Differential Word Problem
Questions 16 to 19 are linked. They refer to the same scenario below.
At a food festival, customers join a popular noodle stall queue at a constant rate of \(12\) customers per minute. The organisers assign more staff when the queue is longer, so customers leave the queue at a rate proportional to the number of customers currently in the queue.
16. Let \(N\) be the number of customers in the queue \(t\) minutes after the stall opens. Which differential equation models \(N\), where \(k>0\)?Tested fundamental: Forming a net-rate differential equation from a linked word problem.
Explanation: Customers join the queue at a constant rate of \(12\) customers per minute. Customers leave the queue at a rate proportional to the current number of customers, so the leaving rate is \(kN\). Therefore, \(\frac{dN}{dt}=12-kN\).
Common wrong choice: A adds the leaving rate instead of subtracting it.
17. The organisers observe that when there are \(60\) customers in the queue, the size of the queue stays unchanged. Find the value of \(k\).
Answer format: Give your answer as an exact fraction, e.g. 2/3.
Answer: \(\displaystyle \frac15\)
Tested fundamental: Using a steady-state condition to find a constant of proportionality.
Explanation: When there are \(60\) customers and the queue size stays unchanged, the rate at which customers join equals the rate at which customers leave. Since customers join at a rate of \(12\) customers per minute and leave at a rate of \(kN\), \(12=k(60)\). Hence \(k=\frac{12}{60}=\frac15\).
Common error: Treating \(60\) as the initial number of customers instead of using it as the queue size where the net change is zero.
18. Initially, there are \(10\) customers in the queue. Using the correct value of \(k\) from Question 17, solve the differential equation. Which expression for \(N\) in terms of \(t\) is obtained?
Tested fundamental: Solving a first-order differential equation from a word problem using separation of variables.
Explanation: From Questions 16 and 17, \(\frac{dN}{dt}=12-\frac15N=\frac{60-N}{5}\). Separating variables gives \(\frac{1}{60-N}\,dN=\frac15\,dt\). Integrating, \(-\ln|60-N|=\frac{t}{5}+C\), so \(60-N=Ae^{-t/5}\).
Using \(N=10\) when \(t=0\), \(60-10=A\), so \(A=50\). Therefore, \(N=60-50e^{-t/5}\).
Common wrong choice: C uses \(10\) as the exponential coefficient instead of applying the initial condition to \(60-N\).
19. Find the time taken for the queue to reach \(50\) customers, giving your answer to \(3\) significant figures.
Answer format: Give your answer in minutes, correct to 3 significant figures.
Answer: \(8.05\)
Tested fundamental: Using the solved model to find the time for a quantity to reach a given value.
Explanation: From Question 18, \(N=60-50e^{-t/5}\). When \(N=50\), \(50=60-50e^{-t/5}\). So \(e^{-t/5}=\frac15\). Taking natural logarithms, \(-\frac{t}{5}=\ln\frac15=-\ln5\). Therefore, \(t=5\ln5\approx8.05\).
Common error: Rounding too early or giving the exact form \(5\ln5\) when the question asks for \(3\) significant figures.
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