Differential equations – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method (integrating factor): write the equation as \(\dfrac{dy}{dx}+Py=Q\), multiply by \(I=e^{\int P\,dx}\); the left-hand side becomes \(\dfrac{d}{dx}(Iy)\). Integrate both sides, remember the constant, then divide by \(I\). Mark scheme as shown in brackets.

   1. \(I=e^{-3\ln x}=x^{-3}\) [1]. \(\dfrac d{dx}(yx^{-3})=x^{-1}\) [1]. \(yx^{-3}=\ln x+C\) [1]. \(y(1)=2\) gives \(C=2\): \(y=x^3(\ln x+2)\) [1]

   Check: \(y'=3x^2(\ln x+2)+x^2\), and \(y'-\dfrac{3y}x=x^2\), as required.

   The integrating factor uses \(P=-\dfrac3x\), not \(+\dfrac3x\): the sign matters.

2. Answer to Question 2 (6 marks)

   Method: the general solution of \(ay''+by'+cy=f(x)\) is the complementary function (solution with \(f=0\)) plus a particular integral, whose form copies \(f(x)\). Use the initial conditions only on the complete general solution. Mark scheme as shown in brackets.

   1. CF: \(m^2-5m+6=0\), \(m=2,3\): \(Ae^{2x}+Be^{3x}\) [1]. PI: \(y=ke^x\) gives \(k(1-5+6)=4\) [1], \(k=2\) [1]. General solution \(y=Ae^{2x}+Be^{3x}+2e^{x}\) [1]
   2. \(A+B+2=5\) and \(2A+3B+2=7\) [1] give \(A+B=3\), \(2A+3B=5\), \(B=-1\), \(A=4\): \(y=4e^{2x}-e^{3x}+2e^{x}\) [1]

3. Answer to Question 3 (4 marks)

   Method: for \(ay''+by'+cy=0\), solve the auxiliary equation \(am^2+bm+c=0\). Two real roots give \(Ae^{m_1x}+Be^{m_2x}\); a repeated root gives \((A+Bx)e^{mx}\); complex roots \(p\pm qi\) give \(e^{px}(A\cos qx+B\sin qx)\). Mark scheme as shown in brackets.

   1. \(m^2+4m+20=0\), \(m=-2\pm4i\) [1]. \(y=e^{-2x}(A\cos4x+B\sin4x)\) [1]. \(y(0)=A=2\) [1]. \(y'(0)=-2A+4B=0\), \(B=1\): \(y=e^{-2x}(2\cos4x+\sin4x)\) [1]

   The motion is a damped oscillation: the factor \(e^{-2x}\) makes the amplitude shrink quickly.

   The roots come from \(m=\dfrac{-4\pm\sqrt{16-80}}2=\dfrac{-4\pm8i}2\).

4. Answer to Question 4 (3 marks)

   Method: \(\ddot x=-\omega^2x\) is simple harmonic motion with period \(\dfrac{2\pi}\omega\), maximum speed \(a\omega\) and \(v^2=\omega^2(a^2-x^2)\). Here \(\omega^2=25\) and \(a=0.2\). Mark scheme as shown in brackets.

   1. \(\omega=5\), period \(\dfrac{2\pi}5=\) 1.26 s [1]
   2. \(v_{max}=\omega a=5\times0.2=\) 1 m s\(^{-1}\) [1]
   3. \(v^2=25(0.04-0.0144)=0.64\), \(v=\) 0.8 m s\(^{-1}\) [1]

   Check (c): at \(x=0.12\), \(\sqrt{a^2-x^2}=\sqrt{0.0256}=0.16\), so \(v=5\times0.16=0.8\) m s\(^{-1}\) ✓. The period does not depend on the amplitude, a key property of simple harmonic motion. The exact period \(\tfrac{2\pi}5\) s is also accepted.

5. Answer to Question 5 (3 marks)

   Method: substitute \(\theta=T-15\), the excess temperature over the garage, so that \(\dfrac{d\theta}{dt}=-0.05\theta\), an exponential decay. Use \(T=75\) at \(t=0\) to find the constant. Mark scheme as shown in brackets.

   1. \(\dfrac d{dt}(T-15)=-0.05(T-15)\) gives \(T-15=Ke^{-0.05t}\) [1]; \(K=60\) [1]
   2. \(30=15+60e^{-0.05t}\), \(e^{-0.05t}=\tfrac14\), \(t=20\ln4=\) 27.7 min [1]

   So the engine cools to \(30^\circ\)C after about 28 minutes; the excess temperature halves every \(20\ln2\approx13.9\) minutes.

   Separating variables, \(\ln|T-15|=-0.05t+c\), gives the same result.

Total : 20 marks