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2014 Paper 3 Q9
D: 1700.0 B: 1500.0
variable force linear air resistance projectile vector equation of motion differential equation exponential decay trigonometric inequality hyperbolic functions verification

A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).

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2013 Paper 1 Q7
D: 1516.0 B: 1516.0
differential equation first order homogeneous ODE substitution separable equation integration y=ux substitution

  1. Use the substitution \(y=ux\), where \(u\) is a function of \(x\), to show that the solution of the differential equation \[ \frac{\d y}{\d x} = \frac x y + \frac y x \quad \quad (x > 0, y> 0) \] that satisfies \(y=2\) when \(x=1\) is \[ y= x\, \sqrt{4+2\ln x \, } ( x > \e^{-2}). \]
  2. Use a substitution to find the solution of the differential equation \[ \frac{\d y}{\d x} = \frac x y + \frac {2y} x \quad \quad (x > 0, y > 0) \] that satisfies \(y=2\) when \(x=1\).
  3. Find the solution of the differential equation \[ \frac{\d y}{\d x} = \frac {x^2} y + \frac {2y} x \quad \quad (x> 0, \ y> 0) \] that satisfies \(y=2\) when \(x=1\).

Solution:

  1. Let \(y = ux\), then \(\frac{\d y}{\d x} = x\frac{\d u}{\d x} = u\) and the differential equation becomes, \begin{align*} && xu' + u &= \frac{1}{u} +u \\ \Rightarrow && u' &= \frac{1}{ux} \\ \Rightarrow && u u' &= \frac1{x} \\ \Rightarrow && \frac12 u^2 &= \ln x + C \\ (x,y) = (1,2): && \frac12 4 &= C \\ \Rightarrow && \frac12 \frac{y^2}{x^2} &= \ln x + 2 \\ \Rightarrow && y^2 &= x^2 \l 2\ln x + 4 \r \\ \Rightarrow && y &= x \sqrt{4 + 2 \ln x} \quad (x > e^{-2}) \end{align*}
  2. Let \(y = ux^2\) then \begin{align*} && \frac{\d y}{\d x} &= \frac{x^2}{y} + \frac{2y}{x} \\ \Rightarrow && u' x^2 + 2x u &= \frac{1}{u} + 2x u \\ \Rightarrow && u' u &= \frac{1}{x^2} \\ \Rightarrow && \frac12 u^2 &= -\frac{1}{x} + C \\ (x,y) = (1,2): && 2 &= C - 1 \\ \Rightarrow && \frac12 \frac{y^2}{x^4} &= 3 - \frac{1}{x} \\ \Rightarrow && y &= x\sqrt{2(3x^2-x)}, \quad (x > \frac13) \end{align*}

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2013 Paper 3 Q2
D: 1700.0 B: 1516.0
Taylor series Maclaurin series Leibniz rule differential equation arcsin recurrence relation infinite series evaluation repeated differentiation

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

Solution: \begin{align*} && y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\ && \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\ &&&= \frac{1+ xy}{1-x^2} \\ \Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\ \frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\ &&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\ \end{align*} Since \(y(0) = 0, y'(0) = 1\) we can look at the recursion: \(y^{(n+2)} - (n+1)^2y^{(n)}\) for larger terms, ie \(y^{(2k)}(0) = 0\) \(y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2\) and \(y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2\). Therefore \begin{align*} && \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\ \\ \Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\ &&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\ &&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\ \Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}} \end{align*}

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2013 Paper 3 Q7
D: 1700.0 B: 1500.0
differential equation second order energy method conservation law differentiation inequality hyperbolic functions Lyapunov function

  1. Let \(y(x)\) be a solution of the differential equation \( \dfrac {\d^2 y}{\d x^2}+y^3=0\) with \(y = 1\) and \(\dfrac{\d y}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d y}{\d x}\right)^{\!\!2} + \tfrac 12 y^4\,. \] Show by differentiation that \({\rm E}\) is constant and deduce that \( \vert y(x) \vert \le 1\) for all \(x\).
  2. Let \(v(x)\) be a solution of the differential equation \( \dfrac{\d^2 v}{\d x^2} + x \dfrac {\d v}{\d x} +\sinh v =0\) with \(v = \ln 3\) and \(\dfrac{\d v}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d v}{\d x}\right)^{\!\!2} + 2 \cosh v\,. \] Show that \(\dfrac{\d{\rm E}}{\d x}\le 0\) for \(x\ge0\) and deduce that \(\cosh v(x) \le \frac53\) for \(x\ge0\).
  3. Let \(w(x)\) be a solution of the differential equation \[ \frac{\d^2 w}{\d x^2} + (5\cosh x - 4 \sinh x -3) \frac{\d w}{\d x} + (w\cosh w + 2 \sinh w) =0 \] with \(\dfrac{\d w }{\d x}=\dfrac 1 { \sqrt 2 }\) and \(w=0\) at \(x=0\). Show that \(\cosh w(x) \le \frac54\) for \(x\ge0\).

Solution:

  1. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ \Rightarrow && E'(x) &= 2 \frac{\d y}{\d x} \frac{\d^2 y}{\d x^2} + 2y^3 \frac{\d y}{\d x} \\ &&&= 2\frac{\d y}{\d x} \left ( \frac{\d^2 y}{\d x^2} + y^3 \right) \\ &&&= 0 \end{align*} Therefore \(E\) is constant. \(E(0) = \frac12\) and \begin{align*} &&\frac12 &= \left ( \frac{\d y}{\d x} \right)^2 + \frac12 y^4 \\ &&&\geq \frac12 y^4 \\ \Rightarrow && |y| &\leq 1 \end{align*}
  2. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ \Rightarrow && E'(x) &= 2 \frac{\d v}{\d x}\frac{\d^2 v}{\d^2 x} + 2 \sinh v \frac{\d v}{\d x} \\ &&&= 2 \frac{\d v}{\d x} \left ( \frac{\d^2 v}{\d^2 x} + \sinh v\right) \\ &&&= 2 \frac{\d v}{\d x} \left ( -x \frac{\d v}{\d x}\right) \\ &&&= -2x \left ( \frac{\d v}{\d x} \right)^2 \leq 0 \tag{\(x \geq 0\)} \\ \\ && E(0) &= 0^2 + 2 \cosh \ln 3 \\ &&&= 3 + \frac13 = \frac{10}{3} \\ \Rightarrow && \frac{10}{3} &\geq E(x) \\ &&&= \left ( \frac{\d v}{\d x} \right)^2 + 2 \cosh v \\ &&&\geq 2 \cosh v \\ \Rightarrow && \cosh v &\leq \frac53 \end{align*}
  3. \(\,\) \begin{align*} && E(x) &= \left ( \frac{\d w}{\d x} \right)^2 + 2(w \sinh w + \cosh w) \\ && E'(x) &= 2 \frac{\d w}{\d x}\frac{\d^2 w}{\d^2 x} + 2(w \cosh w + 2 \sinh w) \frac{\d w}{\d x} \\ &&&= 2 \frac{\d w}{\d x} \left ( \frac{\d^2 w}{\d^2 x}+(w \cosh w + 2 \sinh w)\right) \\ &&&= -2 \left ( \frac{\d w}{\d x} \right)^2 \left (\underbrace{5\cosh x - 4 \sinh x -3}_{\geq0} \right) \\ &&&\leq 0 \\ && E(0) &= \frac12 + 2 = \frac52 \\ \Rightarrow && \frac52 &\geq E(x) \\ &&&=\underbrace{ \left ( \frac{\d w}{\d x} \right)^2}_{\geq0} + 2(\underbrace{w \sinh w}_{\geq 0} + \cosh w) \\ &&&\geq2\cosh w \\ \Rightarrow && \cosh w &\leq \frac54 \end{align*}

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2013 Paper 3 Q9
D: 1700.0 B: 1484.8
differential equation second order SHM small oscillations Archimedes' principle buoyancy sphere volume of revolution equilibrium linearisation period

A sphere of radius \(R\) and uniform density \(\rho_{\text{s}}\) is floating in a large tank of liquid of uniform density \(\rho\). Given that the centre of the sphere is a distance \(x\) above the level of the liquid, where \(x < R\), show that the volume of liquid displaced is \[ \frac \pi 3 (2R^3-3R^2x +x^3)\,. \] The sphere is acted upon by two forces only: its weight and an upward force equal in magnitude to the weight of the liquid it has displaced. Show that \[ 4 R^3\rho_{\text{s}} (g+\ddot x) = (2R^3 -3R^2x +x^3)\rho g\,. \] Given that the sphere is in equilibrium when \(x=\frac12 R\), find \(\rho_{\text{s}}\) in terms of \(\rho\). Find, in terms of \(R\) and \(g\), the period of small oscillations about this equilibrium position.

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2012 Paper 1 Q8
D: 1516.0 B: 1484.0
differential equation first order integrating factor homogeneous ODE substitution separable equation implicit solution

  1. Show that substituting \(y=xv\), where \(v\) is a function of \(x\), in the differential equation \[ xy \frac{\d y}{\d x} +y^2- 2x^2 =0 \quad (x\ne0) \] leads to the differential equation \[ xv\frac{\d v}{\d x} +2v^2 -2=0\,. \] Hence show that the general solution can be written in the form \[ x^2(y^2 -x^2) = C \,,\] where \(C\) is a constant.
  2. Find the general solution of the differential equation \[ y \frac{\d y}{\d x} +6x +5y=0\, \quad (x\ne0). \]

Solution:

  1. \(\,\) \begin{align*} && y &= xv \\ && y' &= v + xv' \\ \Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\ &&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\ \Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\ \\ \Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\ \Rightarrow && \int \frac{v}{1-v^2} \d v &=2 \ln |x| \\ \Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\ \Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\ \Rightarrow && x^4(1-v^2) &= K \\ \Rightarrow && x^2(x^2-y^2) &= K \end{align*}
  2. \(\,\) \begin{align*} && 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\ &&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\ \Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\ \Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\ \Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\ &&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\ \Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\ \Rightarrow && -\ln |x| &= \ln \frac{|v+3|^3}{|v+2|^2} + C \\ \Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\ \Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\ \Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\ \Rightarrow && (y+2x)^2 &= A|y+3x|^3 \end{align*}

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2012 Paper 3 Q1
D: 1700.0 B: 1500.0
differential equation second order substitution chain rule product rule initial value problem nonlinear ODE reduction of order

Given that \(\displaystyle z = y^n \left( \frac{\d y}{\d x}\right)^{\!2}\), show that \[ \frac{\d z}{\d x} = y^{n-1} \frac{\d y}{\d x} \left( n \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2}\right) . \]

  1. Use the above result to show that the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2} = \sqrt y \ \ \ \ \ \ \ \ \ \ (y>0) \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\) is \(y= \big ( \frac38 x^2+1\big)^{\frac23}\).
  2. Find the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} -y \frac{\d^2y}{\d x^2} + y^2=0 \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\).

Solution: \begin{align*} &&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\ \Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\ &&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right) \end{align*}

  1. Let \(z = y (y')^2\), then \begin{align*} && \frac{\d z}{\d x} &= y' \sqrt{y} \\ &&&= \sqrt{z} \\ \Rightarrow && \int z^{-1/2} \d z &= x+C \\ \Rightarrow && 2\sqrt{z} &= x + C \\ x = 0, z=0: && C &= 0 \\ \Rightarrow && y(y')^2 &= \frac14 x^2 \\ \Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\ \Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\ \Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\ x = 0, y = 1: && K &= \frac23 \\ \Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3} \end{align*}
  2. Let \(z = y^{-2} (y')^2\) \begin{align*} && \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\ &&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\ &&&= 2y^{-1}(y') = 2 \sqrt{z} \\ \Rightarrow && 2\sqrt{z} &= 2x + C \\ x = 0, z = 0: && C&= 0 \\ \Rightarrow && z &= x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= xy \\ \Rightarrow && \ln |y| &= \frac12 x^2 + K \\ x =0 , y =1; && K &= 0 \\ \Rightarrow && y &= e^{\frac12 x^2} \end{align*}

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2012 Paper 3 Q7
D: 1700.0 B: 1484.0
differential equation second order auxiliary equation coupled differential equations exponential decay initial conditions geometric series pharmacokinetics superposition

A pain-killing drug is injected into the bloodstream. It then diffuses into the brain, where it is absorbed. The quantities at time \(t\) of the drug in the blood and the brain respectively are \(y(t)\) and \(z(t)\). These satisfy \[ \dot y = - 2(y-z)\,, \ \ \ \ \ \ \ \dot z = - \dot y -3z\, , \] where the dot denotes differentiation with respect to \(t\). Obtain a second order differential equation for \(y\) and hence derive the solution \[ y= A\e^{-t} + B\e ^{-6t}\,, \ \ \ \ \ \ \ z= \tfrac12 A \e^{-t} - 2 B \e^{-6t}\,, \] where \(A\) and \(B\) are arbitrary constants. \begin{questionparts} \item Obtain the solution that satisfies \(z(0)=0\) and \(y(0)= 5\). The quantity of the drug in the brain for this solution is denoted by \(z_1(t)\). \item Obtain the solution that satisfies $ z(0)=z(1)= c$, where \(c\) is a given constant. %\[ %C=2(1-\e^{-1})^{-1} - 2(1-\e^{-6})^{-1}\,. %\] The quantity of the drug in the brain for this solution is denoted by \(z_2(t)\). \item Show that for \(0\le t \le 1\), \[ z_2(t) = \sum _{n=-\infty}^{0} z_1(t-n)\,, \] provided \(c\) takes a particular value that you should find. \end {questionparts}

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2011 Paper 1 Q7
D: 1500.0 B: 1500.0
differential equation first order separable partial fractions logarithmic approximation flow rate modelling integration

In this question, you may assume that \(\ln (1+x) \approx x -\frac12 x^2\) when \(\vert x \vert \) is small. The height of the water in a tank at time \(t\) is \(h\). The initial height of the water is \(H\) and water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.

  1. Suppose that water leaks out at a rate proportional to the height of the water in the tank, and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that \[ \frac {\d h}{\d t } = k( \alpha^2 H -h)\,, \] for some positive constant \(k\). Deduce that the time \(T\) taken for the water to reach height \(\alpha H\) is given by \[ kT = \ln \left(1+\frac1\alpha\right)\,, \] and that \(kT\approx \alpha^{-1}\) for large values of \(\alpha\).
  2. Suppose that the rate at which water leaks out of the tank is proportional to \(\sqrt h\) (instead of \(h\)), and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that the time \(T'\) taken for the water to reach height \(\alpha H\) is given by \[ cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln \left(1+\frac1 {\sqrt\alpha} \right)\right)\, \] for some positive constant \(c\), and that \(cT'\approx \sqrt H\) for large values of \(\alpha\).

Solution:

  1. \begin{align*} \frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}} \end{align*}. We also know that when \(h = \alpha^2 H\), \(\frac{\d h}{\d t} = 0\), ie \(c - k \alpha^2 H = 0\) therefore: \[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \] \begin{align*} && \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\ && \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\ && - \ln |\alpha^2H -h| &= kt + C \\ t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\ \Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\ && kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\ &&&= \ln \frac{1+\alpha}{\alpha} \\ &&&= \ln \left (1 + \frac1{\alpha} \right) \\ &&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\ &&&\approx \alpha^{-1} \end{align*}
  2. \begin{align*} && \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\ \Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\ u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\ &&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}} \right)\right)\\ &&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}} \right)\right)}\\ &&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha} \right)\right) \\ &&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\ &&&= \sqrt{H} \end{align*} as required.

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2011 Paper 2 Q6
D: 1600.0 B: 1500.7
integration integration by parts differential equation trigonometric reduction formula tan sec substitution product rule

For any given function \(\f\), let \[ I = \int [\f'(x)]^2 \,[\f(x)]^n \d x\,, \tag{\(*\)} \] where \(n\) is a positive integer. Show that, if \(\f(x)\) satisfies \(\f''(x) =k \f(x)\f'(x)\) for some constant \(k\), then (\(*\)) can be integrated to obtain an expression for \(I\) in terms of \(\f(x)\), \(\f'(x)\), \(k\) and \(n\).

  1. Verify your result in the case \(\f(x) = \tan x\,\). Hence find \[ \displaystyle \int \frac{\sin^4x}{\cos^{8}x} \, \d x\;. \]
  2. Find \[ \displaystyle \int \sec^2x\, (\sec x + \tan x)^6\,\d x\;. \]

Solution: If \(f''(x) = kf(x)f'(x)\) then we can see \begin{align*} && I &= \int [\f'(x)]^2 \,[\f(x)]^n \d x \\ &&&= \int f'(x) \cdot f'(x) [f(x)]^n \d x \\ &&&= \left[ f'(x) \cdot \frac{[f(x)]^{n+1}}{n+1} \right] - \int f''(x) \frac{[f(x)]^{n+1}}{n+1} \d x \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - \int kf'(x) [f(x)]^{n+2} \d x \right) \\ &&&= \frac{1}{n+1} \left (f'(x) [f(x)]^{n+1} - k \frac{[f(x)]^{n+3}}{n+3} \right) +C\\ &&&= \frac{[f(x)]^{n+1}}{n+1} \left ( f'(x) - \frac{k[f(x)]^2}{n+3} \right) + C \end{align*}

  1. If \(f(x) = \tan x, f'(x) = \sec^2 x, f''(x) = 2 \sec^2 x \tan x = 2 \cdot f(x) \cdot f'(x)\), so \(\tan\) satisfies the conditions for the theorem. \begin{align*} && I &= \int \sec^4 x \tan^n x \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - \int 2 \sec^2 x \tan x \cdot \frac{\tan^{n+1} x}{n+1} \d x \\ &&&= \left [\sec^2x \cdot \frac{\tan^{n+1} x}{n+1} \right] - 2 \cdot \frac{\tan^{n+3} x}{(n+1)(n+3)} \\ \end{align*} So \begin{align*} && I &= \int \frac{\sin^4 x}{\cos^8 x} \d x \\ &&&= \int \tan^4 x \sec^4 x \d x \\ &&&= \int [\sec^2 x]^2 [\tan x]^4 \d x \\ &&&= \frac{\tan^{5}x}{5} \left ( \sec^2 x - \frac{2 \tan^2 x}{7} \right) + C \end{align*}
  2. \begin{align*} && I &= \int \sec^2x\, (\sec x + \tan x)^6\,\d x \\ &&&= \int (\sec x (\sec x + \tan x))^2 \cdot (\sec x + \tan x )^4 \d x \\ &&&= \frac{(\sec x + \tan x)^5}{5} \left ( \sec x (\sec x + \tan x) - \frac{(\sec x + \tan x)^2}{7} \right) + C \end{align*}

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2011 Paper 3 Q1
D: 1700.0 B: 1500.0
differential equation second order first order integrating factor substitution particular integral linear ODE reduction of order

  1. Find the general solution of the differential equation \[ \frac{\d u}{\d x} - \left(\frac { x +2}{x+1}\right)u =0\,. \]
  2. Show that substituting\(y=z\e^{-x}\) (where \(z\) is a function of \(x\)) into the second order differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = 0 \tag{\(*\)} \] leads to a first order differential equation for \(\dfrac{\d z}{\d x}\,\). Find \(z\) and hence show that the general solution of \((*)\) is \[ y= Ax + B\e^{-x}\,, \] where \(A\) and \(B\) are arbitrary constants.
  3. Find the general solution of the differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = (x+1)^2 . \]

Solution:

  1. \begin{align*} && 0 &= \frac{\d u}{\d x} - \left ( \frac{x+2}{x+1} \right)u \\ \Rightarrow && \int \frac1u \d u &= \int 1 + \frac1{x+1} \d x \\ \Rightarrow && \ln |u| &= x + \ln |x+1| + C \\ \Rightarrow && u &= A(x+1)e^x \end{align*}
  2. If \(y = ze^{-x}\), \(y' = (z'-z)e^{-x}\), \(y'' = (z''-2z'+z)e^{-x}\) \begin{align*} && 0 &= (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y \\ y = ze^{-x}: && 0 &= (x+1) \left ( \frac{\d^2 z}{\d x^2} - 2\frac{\d z}{\d x} +z\right)e^{-x} +x \left ( \frac{\d z}{\d x} -z\right)e^{-x} - ze^{-x} \\ &&&= (x+1) \frac{\d^2 z}{\d x^2} -(x+2)\frac{\d z}{\d x} \\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{\d z}{\d x}\right) &= \left ( \frac{x+2}{x+1}\right) \frac{\d z}{\d x} \end{align*} Therefore \(\frac{\d z}{\d x} = A(x+1)e^x \) and so \begin{align*} z &= A \int (x+1)e^{x} \d x \\ &= A \left ( \left [ (x+1)e^x\right] - \int e^x \d x \right) \\ &= A(x+1)e^x - Ae^x + B \\ y &= Ax + Be^{-x} \end{align*}
  3. We have found the complementary solution. To find a particular integral consider \(y = ax^2 + bx + c\), then \(y' = 2ax+b, y'' = 2a\) and we have \begin{align*} && x^2+2x+1 &= 2a(x+1) + x(2ax+b) - (ax^2+bx+c) \\ \Rightarrow && x^2+2x+1 &= ax^2+ 2ax + 2a-c \\ \Rightarrow && a = 1, &c=1 \end{align*} so the general solution should be \[ y = Ax + Be^{-x} + x^2+1 \]

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2010 Paper 1 Q6
D: 1500.0 B: 1516.0
differential equation second order reduction of order substitution first order integrating factor verification exponential function

Show that, if \(y=\e^x\), then \[ (x-1) \frac{\d^2 y}{\d x^2} -x \frac{\d y}{\d x} +y=0\,. \tag{\(*\)} \] In order to find other solutions of this differential equation, now let \(y=u\e^x\), where \(u\) is a function of \(x\). By substituting this into \((*)\), show that \[ (x-1) \frac{\d^2 u}{\d x^2} + (x-2) \frac{\d u}{\d x} =0\,. \tag{\(**\)} \] By setting \( \dfrac {\d u}{\d x}= v\) in \((**)\) and solving the resulting first order differential equation for \(v\), find \(u\) in terms of \(x\). Hence show that \(y=Ax + B\e^x\) satisfies \((*)\), where \(A\) and \(B\) are any constants.

Solution: \begin{align*} && y &= e^x \\ && y' &= e^x \\ && y'' &= e^x \\ \Rightarrow && (x-1)y'' - x y' + y &= (x-1)e^x - xe^x + e^x \\ &&&= 0 \end{align*} Suppose \(y = ue^x\) then \begin{align*} && y' &= u'e^x + ue^x \\ && y'' &= (u''+u')e^x + (u'+u)e^x \\ &&&= (u''+2u' +u)e^x \\ \\ && 0 &= (x-1)y'' - x y' + y \\ &&&= [(x-1)(u''+2u'+u) - x(u'+u)+u]e^x \\ &&&= [(x-1)u'' +(x-2)u']e^x \\ \Rightarrow && 0 &= (x-1)u'' + (x-2)u' \\ v = u': && 0 &= (x-1)v' + (x-2) v \\ \Rightarrow && \frac{v'}{v} &= -\frac{x-2}{x-1} \\ &&&= -1-\frac{1}{x-1} \\ \Rightarrow && \ln v &= -x - \ln(x-1) + C \\ \Rightarrow && v &= A(x-1)e^{-x} \\ && u &= \int Axe^{-x} - Ae^{-x} \d x \\ &&&= \left [-Axe^{-x} +Ae^{-x} \right] + \int Ae^{-x} \d x \\ &&&= -Axe^{-x} + D\\ \Rightarrow && y &= ue^x \\ &&&= -Ax + De^x \end{align*}

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2010 Paper 3 Q7
D: 1700.0 B: 1516.0
differential equation Maclaurin series proof by induction Leibniz rule recurrence relation Chebyshev polynomial trigonometric higher derivatives

Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\), \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\), \(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.

Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)

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2010 Paper 3 Q8
D: 1700.0 B: 1531.5
integration differential equation first order linear ODE partial fractions quotient rule trigonometric integrating factor rational function

Given that \({\rm P} (x) = {\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)\), write down an expression for \[ \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x\, . \]

  1. By choosing the function \({\rm R}(x)\) to be of the form \(a +bx+c x^2\), find \[ \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x \,.\] Show that the choice of \({\rm R}(x)\) is not unique and, by comparing the two functions \({\rm R}(x)\) corresponding to two different values of \(a\), explain how the different choices are related.
  2. Find the general solution of \[ (1+\cos x +2 \sin x) \frac {\d y}{\d x} +(\sin x -2 \cos x)y = 5 - 3 \cos x + 4 \sin x\,. \]

Solution: \begin{align*} && \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\ &&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\ &&&= \frac{R(x)}{Q(x)} + C \end{align*}

  1. Suppose \(Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3\), and \(R(x) = a +bx + cx^2\), then \begin{align*} && 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\ &&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\ &&&\quad+(2c+2b-6a-2b)x + (b-2a) \\ \Rightarrow && 2c-3b &= 5 \\ && 2c-6a &= -4 \\ && b - 2a &= -3 \\ \Rightarrow && b &= 2a - 3\\ && c &= 3a-2 \end{align*} So say \(a = 0, b = -3, c = -2\) we will have \begin{align*} \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C \end{align*} Suppose we have a different value of \(a\), then we end up with: \begin{align*} \frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2} \end{align*} So the different antiderivatives differ by a constant.
  2. \(\,\) \begin{align*} && \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\ \end{align*} We want to find \(R(x) = A \cos x + B\sin x + C\) such that \begin{align*} &&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\ &&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\ &&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\ &&&=(-A+C) \sin x + (B-2C)\cos x +\\ &&&\quad\quad (2B-A+A-2B)\sin x \cos x \\ &&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\ &&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\ \Rightarrow && B-2A &= 5\\ && C-A &= 4 \\ && B-2C &= -3 \\ \end{align*} There are many solutions so WLOG \(C=4, A = 0, B = 5\) and so \begin{align*} && \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\ \Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x) \end{align*}

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2009 Paper 3 Q11
D: 1700.0 B: 1500.1
variable force momentum variable mass Newton's second law differential equation conservation of momentum dust cloud comet separable ODE limiting behaviour

A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude \(M\!f\) acting in the direction of its motion. When it entered the cloud, the comet had mass \(M\) and speed \(V\). After a time \(t\), it has travelled a distance \(x\) through the cloud, its mass is \(M(1+bx)\), where~\(b\) is a positive constant, and its speed is \(v\).

  1. In the case when \(f=0\), write down an equation relating \(V\), \(x\), \(v\) and \(b\). Hence find an expression for \(x\) in terms of \(b\), \(V\) and \(t\).
  2. In the case when \(f\) is a non-zero constant, use Newton's second law in the form \[ \text{force} = \text{rate of change of momentum} \] to show that \[ v = \frac{ft+V}{1+bx}\,. \] Hence find an expression for \(x\) in terms of \(b\), \(V\), \(f\) and \(t\). Show that it is possible, if \(b\), \(V\) and \(f\) are suitably chosen, for the comet to move with constant speed. Show also that, if the comet does not move with constant speed, its speed tends to a constant as \(t\to\infty\).

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