Differential Equations: exam review Exam review: second paper Worked solutions

The number \(I\) of people infected with a virus in a town of 1200 people changes with time \(t\) weeks. The rate of increase of \(I\) is proportional to the product of the number of infected people and the number who are not infected. [5]

1. (a) (a) Write down a differential equation for \(I\) in terms of \(t\) and a constant \(k\). [2]

   • M1 for a product of \(I\) and \(1200-I\)
   • A1 for \(\frac{dI}{dt}=kI(1200-I)\)

2. (b) (b) A curve satisfies \(\frac{dy}{dx}=8 x - \frac{3}{x^{2}}\) and passes through the point \((2,5)\). Find \(y\) in terms of \(x\). [3]

   • M1 for integrating, at least one term correct
   • A1 for \(4 x^{2} + \frac{3}{x}+c\)
   • A1 for \(c=- \frac{25}{2}\) and \(y=4 x^{2} - \frac{25}{2} + \frac{3}{x}\)

A cell culture is modelled by a differential equation for \(y\), the mass in milligrams, after \(x\) days. [6]

1. (a) (a) Find the general solution of \(\frac{dy}{dx}=3 x^{2}e^{-y}\), giving your answer in the form \(e^{y}=f(x)\). [3]

   • M1 for separating: \(\int e^{y}\,dy=\int 3 x^{2}\,dx\)
   • A1 for \(e^{y}\) and \(x^{3}\)
   • A1 for \(e^{y}=x^{3}+c\)

2. (b) (b) Given that \(y=2\) when \(x=0\), find \(y\) in terms of \(x\). [2]

   • M1 for substituting \(x=0,\ y=2\): \(c=e^{2}\)
   • A1 for \(y=\ln\left(x^{3}+e^{2}\right)\)

3. (c) (c) Find the value of \(y\) when \(x=1\). [1]

   • A1 for \(y=2.13\)

A sample of a radioactive isotope has mass \(m\) mg after \(t\) days, and \(\frac{dm}{dt}=-km\), where \(k\) is a positive constant. At \(t=0\) the mass is 480 mg and it is decreasing at 36 mg per day. [7]

1. (a) (a) Find the value of \(k\). [2]

   • M1 for \(-36=-k\times 480\) (rate \(-36\) when \(m=480\))
   • A1 for \(k=\frac{3}{40}\)

2. (b) (b) Solve the differential equation to show that \(m=480 e^{-kt}\). [2]

   • M1 for \(\int\frac{1}{m}\,dm=\int -k\,dt\), giving \(\ln m=-kt+c\)
   • A1 for using \(m=480\) at \(t=0\) to get \(m=480 e^{-kt}\)

3. (c) (c) Find the half-life. [2]

   • M1 for \(480 e^{-0.0750t}=240\) or \(e^{-kt}=0.5\)
   • A1 for \(t=\frac{\ln 2}{0.0750}=9.24\)

4. (d) (d) Find the time taken for the amount to fall to 15 mg. [1]

   • A1 for \(t=\frac{\ln(32)}{0.0750}=46.2\)

A bowl of soup at \(85^\circ\)C is left in a kitchen at \(25^\circ\)C. Its temperature \(T^\circ\)C after \(t\) minutes satisfies \(\frac{dT}{dt}=-k(T-25)\), where \(k\) is a positive constant. [5]

1. (a) (a) Solve the differential equation, using \(T(0)=85\) and \(T(4)=65\), to find \(T\) in terms of \(t\) and the value of \(k\). [3]

   • M1 for \(\int\frac{1}{T-25}\,dT=\int -k\,dt\), giving \(\ln(T-25)=-kt+c\)
   • A1 for \(T=25+60e^{-kt}\)
   • A1 for \(40=60e^{-4k}\), so \(k=\frac{\ln\frac{60}{40}}{4}=0.101\)

2. (b) (b) Find the time taken for the soup to cool to 35\(^\circ\)C. [2]

   • M1 for \(10=60e^{-0.101t}\)
   • A1 for \(t=\frac{\ln\frac{60}{10}}{k}=17.7\)

A tank holds 500 litres of water in which 30 kg of sugar is dissolved. Syrup containing 0.3 kg of sugar per litre flows in at 5 litres per minute. The mixture is kept well stirred and flows out at 5 litres per minute, so the volume stays at 500 litres. Let \(m\) kg be the mass of sugar in the tank after \(t\) minutes. [6]

1. (a) (a) Show that \(\frac{dm}{dt}=\frac{3}{2}-\frac{m}{100}\). [2]

   • M1 for rate in \(=0.3\times5=\frac{3}{2}\) kg per minute or rate out \(=\frac{m}{500}\times5\)
   • A1 for rate in minus rate out, \(\frac{dm}{dt}=\frac{3}{2}-\frac{m}{100}\) (AG)

2. (b) (b) Solve the differential equation to find \(m\) in terms of \(t\). [2]

   • M1 for \(\int\frac{1}{150-m}\,dm=\int\frac{1}{100}\,dt\), giving \(-\ln(150-m)=\frac{t}{100}+c\)
   • A1 for \(m=150-120e^{-t/100}\) (using \(m(0)=30\))

3. (c) (c) Find the time taken for the mass of sugar to reach 100 kg. [1]

   • A1 for \(50=120e^{-t/100}\), so \(t=100\ln\frac{120}{50}=87.5\)

4. (d) (d) Write down the mass of sugar in the tank in the long run. [1]

   • A1 for \(m\to150\) as \(t\to\infty\)

The slope field shows the differential equation \(\frac{dy}{dx}=\frac{x(y+1)}{2}\) for integer values of \(x\) and \(y\). [6]

1. (a) (a) Find the gradient of the solution curve at the points \((2,3)\) and \((-1,-3)\). [2]

   • M1 for substituting \(x,y\) into the right side
   • A1 for \(4\) and \(1\)

2. (b) (b) Show that \(y=-1\) is a constant solution of the differential equation. [1]

   • R1 for \(y=-1\) gives \(\frac{dy}{dx}=0\) for every \(x\)

3. (c) (c) Copy the slope field and sketch the solution curve that passes through \((0,0)\). [2]

   • A1 for a curve through \((0,0)\) following the segments
   • A1 for a minimum at \((0,0)\) and the curve rising on both sides, above \(y=-1\)

4. (d) (d) Use your curve to estimate the value of \(y\) when \(x=2\). [1]

   • A1 for an estimate close to \(1.7\) (the exact solution gives \(1.72\))

The population \(P\) (in hundreds) of deer in a reserve satisfies \(\frac{dP}{dt}=0.3P\left(1-\frac{P}{80}\right)\), where \(t\) is the time in years and \(P>0\). [6]

1. (a) (a) Find the two values of \(P\) for which \(\frac{dP}{dt}=0\). [2]

   • M1 for setting \(\frac{3}{10}P\left(1-\frac{P}{80}\right)=0\)
   • A1 for \(P=0\) and \(P=80\)

2. (b) (b) By finding the sign of \(\frac{dP}{dt}\) when \(P=70\) and when \(P=90\), decide which of the two values in part (a) is stable. Give a reason. [2]

   • M1 for \(P=70\): \(\frac{dP}{dt}=2.62>0\) and \(P=90\): \(\frac{dP}{dt}=-3.38<0\)
   • R1 for solutions either side move towards \(P=80\), so \(P=80\) is stable

3. (c) (c) Write down the value of \(P\) for which the population is growing fastest. [1]

   • A1 for \(P=\frac{80}{2}=40\)

4. (d) (d) Find the greatest rate of growth. [1]

   • A1 for \(\frac{3}{10}\times40\times\frac{1}{2}=6\)

The variables \(x\) and \(y\) measure the angle and the angular speed of a swing, as differences from their rest values. They satisfy the system \(\frac{dx}{dt}=- 3 x + 2 y,\quad \frac{dy}{dt}=- 4 x + y\). [6]

1. (a) (a) Find the eigenvalues of the matrix of the system. [2]

   • M1 for \(\lambda^{2}+2\lambda+5=0\) (trace and determinant)
   • A1 for \(\lambda=-1\pm 2 i\)

2. (b) (b) State the type of the equilibrium point at the origin, and describe what happens to \(x\) and \(y\) in the long run. [2]

   • A1 for stable focus (real part \(-1\) is negative)
   • A1 for a spiral in: \(x\to0\) and \(y\to0\) (oscillating)

3. (c) (c) Find \(\frac{dx}{dt}\) at the point \((0,1)\), and hence state whether the spiral is clockwise or anticlockwise. [2]

   • M1 for substituting \(x=0,\ y=1\): \(\frac{dx}{dt}=2\)
   • A1 for x increasing on the positive \(y\)-axis, so the spiral is clockwise

A swing moves so that its angle \(\theta\) radians from the vertical after \(t\) seconds satisfies \(\frac{d^{2}\theta}{dt^{2}}+8\sin\theta=0\), with \(\theta(0)=0.5\) and \(\theta'(0)=0.3\). [5]

1. (a) (a) Let \(\omega=\frac{d\theta}{dt}\). Write the equation as a pair of first order equations and use Euler's method with step length \(0.1\) to estimate \(\theta(0.2)\). Work in radians. [3]

   • A1 for \(\frac{d\theta}{dt}=\omega,\ \frac{d\omega}{dt}=-8\sin\theta\)
   • M1 for \(\theta_{n+1}=\theta_n+0.1\omega_n,\ \omega_{n+1}=\omega_n-0.1\times8\sin\theta_n\)
   • A1 for \(\theta_1=0.5300,\ \omega_1=-0.0835\) and \(\theta(0.2)\approx0.5216\)

2. (b) (b) A simpler model replaces \(\sin\theta\) by \(\theta\). Use Euler's method with the same step length to estimate \(\theta(0.2)\) for this model. [2]

   • M1 for \(\omega_{n+1}=\omega_n-0.1\times8\theta_n\), giving \(\theta_1=0.5300,\ \omega_1=-0.1000\)
   • A1 for \(\theta(0.2)\approx0.5200\)

A biologist models the growth of a culture by \(\frac{dy}{dx}=3xy^{2}\), where \(y\) is the mass in grams after \(x\) days, and \(y(0)=0.25\). [14]

1. (a) (a) Use Euler's method with step length \(0.5\) to estimate \(y(1)\). [3]

   • M1 for \(y_{n+1}=y_n+0.5(3x_ny_n^{2})\)
   • A1 for \(y_1=0.2500\)
   • A1 for \(y(1)\approx 0.297\)

2. (b) (b) Use Euler's method with step length \(0.25\) to estimate \(y(1)\). [3]

   • M1 for \(y_{n+1}=y_n+0.25(3x_ny_n^{2})\) used with \(4\) steps
   • A1 for \(y_{1}=0.2500\), \(y_{2}=0.2617\)
   • A1 for \(y(1)\approx 0.334\)

3. (c) (c) Solve the differential equation, giving \(y\) in terms of \(x\). [4]

   • M1 for separating: \(\int\frac{dy}{y^{2}}=\int 3x\,dx\)
   • A1 for \(-\frac{1}{y}=\frac{3 x^{2}}{2}+c\)
   • A1 for \(c=-4\) from \(y(0)=0.25\)
   • A1 for \(y=\frac{1}{4 - \frac{3 x^{2}}{2}}\)

4. (d) (d) Find the percentage error in your estimate from part (b). [2]

   • M1 for \(\frac{|0.4000-0.3339|}{0.4000}\times100\) (exact \(y(1)=0.4000\))
   • A1 for \(16.5\%\)

5. (e) (e) Explain why both estimates are smaller than the exact value, and say how to get a more accurate estimate. [2]

   • R1 for the solution curve is concave up (its gradient increases), so each tangent line lies below the curve
   • R1 for a smaller step length (more steps) gives a more accurate estimate

The numbers of plankton \(x\) and krill \(y\) in a sample of sea water, \(t\) days after the first count, are modelled by \(\frac{dx}{dt}=0.5x-0.02xy,\quad \frac{dy}{dt}=0.005xy-0.4y\), with \(x(0)=120\) and \(y(0)=30\). [14]

1. (a) (a) Write down the Euler recurrence relations for \(x_{n+1}\) and \(y_{n+1}\) with step length \(h\). [2]

   • M1 for \(x_{n+1}=x_n+h(0.5x_n-0.02x_ny_n)\)
   • A1 for \(y_{n+1}=y_n+h(0.005x_ny_n-0.4y_n)\) (old \(x\) and \(y\) in both)

2. (b) (b) Use Euler's method with \(h=0.5\) to estimate \(x(1)\) and \(y(1)\). [4]

   • M1 for substituting \(x_0=120,\ y_0=30\) with \(h=0.5\)
   • A1 for \(x_1=114.0,\ y_1=33.00\)
   • M1 for a second step using \(x_1,\ y_1\)
   • A1 for \(x(1)\approx105,\ y(1)\approx35.8\)

3. (c) (c) Use technology and Euler's method with \(h=0.1\) to estimate \(x(4)\) and \(y(4)\). [3]

   • M1 for a table of \(t,x,y\) with \(h=0.1\) (40 steps)
   • A1 for \(x(4)\approx51.8\)
   • A1 for \(y(4)\approx32.5\)

4. (d) (d) Use technology and Euler's method with \(h=0.1\) to find the largest value of \(y\) for \(0\leq t\leq10\). [3]

   • M1 for a table or graph of \(y\) for \(0\leq t\leq10\)
   • A1 for the largest value at \(t=2.1\)
   • A1 for \(37.8\)

5. (e) (e) Euler's method with \(h=0.1\) gives successive peaks of \(x\) of \(132\) (at \(t=13.6\)) and \(142\) (at \(t=28.1\)). A student says that the model predicts that the peaks of \(x\) will keep growing. Comment on the student's statement. [2]

   • A1 for the peaks of \(x\) increase (\(132\) then \(142\))
   • R1 for the growth is caused by the error in Euler's method: the exact solution of this model is a repeating cycle whose peaks do not grow, so the statement is not supported (a smaller step length reduces the growth)

Two strains of bacteria in a dish feed on each other's waste. Their populations \(x\) and \(y\) in thousands, after \(t\) hours, satisfy \(\frac{dx}{dt}=4 x + 3 y,\quad \frac{dy}{dt}=x + 2 y\), with \(x(0)=22\) and \(y(0)=6\). [16]

1. (a) (a) Write the system in the form \(\frac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}=A\begin{pmatrix}x\\y\end{pmatrix}\), stating \(A\). [1]

   • A1 for \(A=\begin{pmatrix}4&3\\1&2\end{pmatrix}\)

2. (b) (b) Find the eigenvalues of \(A\). [2]

   • M1 for \(\lambda^{2}-6\lambda+5=0\) (trace and determinant)
   • A1 for \(\lambda=1,\ 5\)

3. (c) (c) Find an eigenvector for each eigenvalue. [3]

   • M1 for solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\)
   • A1 for \(\lambda=1:\ \begin{pmatrix}1\\-1\end{pmatrix}\)
   • A1 for \(\lambda=5:\ \begin{pmatrix}3\\1\end{pmatrix}\)

4. (d) (d) Write down the general solution of the system. [2]

   • M1 for one correct term \(Ce^{\lambda t}\mathbf{v}\)
   • A1 for \(\begin{pmatrix}x\\y\end{pmatrix}=C\begin{pmatrix}1\\-1\end{pmatrix}e^{t}+D\begin{pmatrix}3\\1\end{pmatrix}e^{5 t}\)

5. (e) (e) Find \(x(t)\) and \(y(t)\). [3]

   • M1 for \(22=1C+3D,\ 6=-1C+1D\)
   • A1 for \(C=1,\ D=7\)
   • A1 for \(x=21 e^{5 t} + e^{t}\) and \(y=7 e^{5 t} - e^{t}\)

6. (f) (f) State the type of the equilibrium point at the origin, and sketch the phase portrait, marking the directions of the eigenvectors. [3]

   • A1 for unstable node (both eigenvalues positive)
   • A1 for straight-line trajectories along the lines through \((1,-1)\) and \((3,1)\)
   • A1 for arrows pointing away from the origin along every trajectory

7. (g) (g) Find the ratio \(y:x\) as \(t\) becomes large. [2]

   • M1 for the term with the larger eigenvalue \(e^{5t}\) dominates
   • A1 for \(y:x=1:3\) (the eigenvector of \(\lambda=5\))

The displacement \(x\) cm of the arm of a door closer \(t\) seconds after release satisfies \(\frac{d^{2}x}{dt^{2}}+5\frac{dx}{dt}+6x=0\), with \(x(0)=3\) and \(x'(0)=-4\). [16]

1. (a) (a) Let \(y=\frac{dx}{dt}\). Write the equation as a pair of first order equations and give the matrix of the system. [2]

   • A1 for \(\frac{dx}{dt}=y\) and \(\frac{dy}{dt}=- 6 x - 5 y\)
   • A1 for the matrix \(\begin{pmatrix}0&1\\-6&-5\end{pmatrix}\)

2. (b) (b) Use Euler's method with step length \(0.25\) to estimate \(x(0.5)\). [3]

   • M1 for \(x_{n+1}=x_n+0.25y_n,\ y_{n+1}=y_n+0.25(- 6 x_{n} - 5 y_{n})\)
   • A1 for \(x_1=2.000,\ y_1=-3.500\)
   • A1 for \(x(0.5)\approx 1.125\)

3. (c) (c) Find the eigenvalues of the matrix, and hence decide whether the solution is stable. [3]

   • M1 for \(\lambda^{2}+5\lambda+6=0\)
   • A1 for \(\lambda=-3\text{ or }-2\)
   • R1 for both eigenvalues negative, so the solution is stable

4. (d) (d) Find the exact solution \(x(t)\). [4]

   • M1 for \(x=Ae^{- 3 t}+Be^{- 2 t}\)
   • M1 for \(A+B=3\) and \(- 3 A - 2 B=-4\)
   • A1 for \(A=-2,\ B=5\)
   • A1 for \(x=5 e^{- 2 t} - 2 e^{- 3 t}\)

5. (e) (e) Find the percentage error of your estimate from part (b). [2]

   • M1 for \(\frac{|1.393-1.125|}{1.393}\times100\) (exact \(x(0.5)=1.393\))
   • A1 for \(19.2\%\)

6. (f) (f) Decide whether the solutions of (i) \(\frac{d^{2}x}{dt^{2}}+3\frac{dx}{dt}-10x=0\) and (ii) \(\frac{d^{2}x}{dt^{2}}-2\frac{dx}{dt}+5x=0\) are stable. [2]

   • A1 for (i) \(\lambda=-5\text{ or }2\), one eigenvalue is positive (a saddle): not stable
   • A1 for (ii) \(\lambda=1\pm 2i\), real part positive: not stable

Two reservoirs are linked by a channel. The volumes \(x\) and \(y\) (in millions of litres) \(t\) days after a storm satisfy \(\frac{dx}{dt}=-2x+2y+4,\ \frac{dy}{dt}=x-4y+1\). [8]

1. (a) (a) Find the coordinates of the equilibrium point. [2]

   • M1 for setting both \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}=0\)
   • A1 for \((3,1)\)

2. (b) (b) Find the eigenvalues of the matrix of coefficients, and hence classify the equilibrium point as stable or unstable. [3]

   • M1 for \(\lambda^{2}+6\lambda+6=0\)
   • A1 for \(\lambda=-4.732\) or \(-1.268\)
   • R1 for both eigenvalues negative, so the equilibrium point is stable

3. (c) (c) A manager claims that \((4,1)\) is an equilibrium point. Use one step of Euler's method with step length \(0.1\) from \((4,1)\) to test the claim. [2]

   • M1 for \(x_1=4+0.1\times(-2),\ y_1=1+0.1\times(1)\)
   • R1 for \((3.80,1.10)\) is not \((4,1)\), so it is not an equilibrium point

4. (d) (d) Write down the long-run values of \(x\) and \(y\) for any starting point. [1]

   • A1 for \(x\to3,\ y\to1\)