Differential equations

Showing 1-25 of 35 problems
2024 Paper 3 Q6
D: 1500.0 B: 1500.0

  1. A particle moves in two-dimensional space. Its position is given by coordinates \((x, y)\) which satisfy \[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\] where \(t\) is the time and \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0)\).
    1. By considering \(\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}\), show that if the particle is at the origin \((0,0)\) at some time \(t > 0\), then it is necessary that \(x_0 = y_0\).
    2. Given that \(x_0 = y_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
  2. A particle whose position in three-dimensional space is given by co-ordinates \((x, y, z)\) moves with time \(t\) such that \[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\] \[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\] where \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0,\, z_0)\).
    1. Show that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(y_0\) is the mean of \(x_0\) and \(z_0\).
    2. Show further that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(x_0 = y_0 = z_0\).
    3. Given that \(x_0 = y_0 = z_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).

Show Solution
    1. \(\,\) \begin{align*} && \frac{\d x}{\d t} - \frac{\d y}{\d t} &= -2x + 2y \\ &&&= -2(x-y) \\ \Rightarrow && \dot{z} &= -2z \tag{\(z = x-y\)} \\ \Rightarrow && z &= Ae^{-2t} \\ z = 0, t : && A &= 0 \\ \Rightarrow && z &= 0 \quad \forall t \\ \Rightarrow && x &= y \quad \forall t \\ \Rightarrow && x_0 &= y_0 \end{align*}
    2. Since \(x = y\) for all \(t\) our equation can we written as \(\frac{\d x}{\d t} = 2x + u\). This has solution \(x = Ae^{2t} - \frac{u}{2}\) we also have \begin{align*} t = 0: && x_0 &= A - \frac{u}{2} \\ t = T: && 0 &= Ae^{2T} - \frac{u}{2} \\ \Rightarrow && x_0 &= \frac{u}{2} - e^{-2T}\frac{u}{2} \\ \Rightarrow && u &= \frac{2x_0}{1-e^{-2T}} \end{align*}
    1. Let \(w = y - \frac12(x+z)\) then \begin{align*} && \dot{w} &= \dot{y} - \tfrac12(\dot{x}+\dot{z}) \\ &&&= (x-2z+u) - \tfrac12(4y-5z+u+x-2y+u) \\ &&&= -y + \tfrac12(x+z) \\ &&&= -w \\ \Rightarrow && w &= Ae^{-t} \\ \text{at origin}: && w &= 0 \\ \Rightarrow && y &= \tfrac12(x+z) \end{align*}
    2. We now have \(\dot{x} = 2x+2z-5z+u = 2x-3z+u\) and \(\dot{z} = x - x- z +u = - z+u\) so in particular \((x - z)' = 2(x-z)\) or \(x-z = Ae^{2t}\) and since we hit the origin \(x = z\) for all \(t\) and so \(y = x = z\) for all \(t\).
    3. Notice we now have \(\dot{x} = -x + u\) or \(x = Ae^{-t} + u\) \begin{align*} t = 0: && x_0 &= A + u \\ t = T: && 0 &= Ae^{-T} + u \\ \Rightarrow && x_0 &= u-e^{T} u \\ \Rightarrow && u &= \frac{x_0}{1-e^{T}} \end{align*}
2021 Paper 2 Q5
D: 1500.0 B: 1500.0

  1. Use the substitution \(y = (x - a)u\), where \(u\) is a function of \(x\), to solve the differential equation \[ (x - a)\frac{dy}{dx} = y - x, \] where \(a\) is a constant.
  2. The curve \(C\) with equation \(y = f(x)\) has the property that, for all values of \(t\) except \(t = 1\), the tangent at the point \(\bigl(t,\, f(t)\bigr)\) passes through the point \((1, t)\).
    1. Given that \(f(0) = 0\), find \(f(x)\) for \(x < 1\). Sketch \(C\) for \(x < 1\). You should find the coordinates of any stationary points and consider the gradient of \(C\) as \(x \to 1\). You may assume that \(z\ln|z| \to 0\) as \(z \to 0\).
    2. Given that \(f(2) = 2\), sketch \(C\) for \(x > 1\), giving the coordinates of any stationary points.

2020 Paper 2 Q2
D: 1500.0 B: 1500.0

The curves \(C_1\) and \(C_2\) both satisfy the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{kxy - y}{x - kxy},\] where \(k = \ln 2\). All points on \(C_1\) have positive \(x\) and \(y\) co-ordinates and \(C_1\) passes through \((1,\,1)\). All points on \(C_2\) have negative \(x\) and \(y\) co-ordinates and \(C_2\) passes through \((-1,\,-1)\).

  1. Show that the equation of \(C_1\) can be written as \((x-y)^2 = (x+y)^2 - 2^{x+y}\). Determine a similar result for curve \(C_2\). Hence show that \(y = x\) is a line of symmetry of each curve.
  2. Sketch on the same axes the curves \(y = x^2\) and \(y = 2^x\), for \(x \geqslant 0\). Hence show that \(C_1\) lies between the lines \(x + y = 2\) and \(x + y = 4\). Sketch curve \(C_1\).
  3. Sketch curve \(C_2\).

2019 Paper 2 Q6
D: 1500.0 B: 1500.0

Note: You may assume that if the functions \(y_1(x)\) and \(y_2(x)\) both satisfy one of the differential equations in this question, then the curves \(y = y_1(x)\) and \(y = y_2(x)\) do not intersect.

  1. Find the solution of the differential equation $$\frac{dy}{dx} = y + x + 1$$ that has the form \(y = mx + c\), where \(m\) and \(c\) are constants. Let \(y_3(x)\) be the solution of this differential equation with \(y_3(0) = k\). Show that any stationary point on the curve \(y = y_3(x)\) lies on the line \(y = -x - 1\). Deduce that solution curves with \(k < -2\) cannot have any stationary points. Show further that any stationary point on the solution curve is a local minimum. Use the substitution \(Y = y + x\) to solve the differential equation, and sketch, on the same axes, the solutions with \(k = 0\), \(k = -2\) and \(k = -3\).
  2. Find the two solutions of the differential equation $$\frac{dy}{dx} = x^2 + y^2 - 2xy - 4x + 4y + 3$$ that have the form \(y = mx + c\). Let \(y_4(x)\) be the solution of this differential equation with \(y_4(0) = -2\). (Do not attempt to find this solution.) Show that any stationary point on the curve \(y = y_4(x)\) lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines? Sketch the curve \(y = y_4(x)\). You should include on your sketch the two straight line solutions and the two lines of stationary points.

Show Solution
  1. Looking for solution of the form \(y = mx+c\) we find that \(m = mx+c+x+1 \Rightarrow m = -1, c = -2\). At stationary points \(\frac{\d y}{\d x} = 0 \Rightarrow y = -x-1\). If \(y_3(0)= k < -2\) then the solution curve lies below \(y = -x-2\) and therefore it cannot cross \(y = -x -2\) to reach \(y = -x-1\) for a stationary point. Suppose \(Y = y+x\) then \(\frac{\d Y}{\d x} = \frac{\d y}{\d x} + 1=Y+2 \Rightarrow Y = Ae^x-2 \Rightarrow y= (k+2)e^x-x-2\)
    TikZ diagram
  2. \(\,\) \begin{align*} && m &= x^2 + (mx+c)^2 -2x(mx+c) - 4x+4(mx+c) + 3 \\ &&0&= (m^2-2m+1)x^2+(2mc-2c-4+4m)x + (c^2+4c+3-m)\\ \Rightarrow && m &= 1 \\ \Rightarrow && 0 &= c^2+4c+2 \\ \Rightarrow &&&= (c+2)^2-2 \\ \Rightarrow && c &= -2 \pm \sqrt{2} \end{align*} Therefore the lines are \(y = x -2-\sqrt{2}\) and \(y = x -2+\sqrt{2}\). Any stationary point will satisfy \(y' = 0\), ie \(0 = x^2+y^2-2xy-4x+4y+3 = (x-y)^2-4(x-y)+3 = (x-y-3)(x-y-1)\) therefore they must lie on \(y = x-1\) or \(y = x-3\). Any point between these lines must have negative gradient (since one factor is positive and one factor is negative).
    TikZ diagram
2016 Paper 1 Q4
D: 1516.0 B: 1502.9

  1. Differentiate $\displaystyle \; \frac z {(1+z^2)^{\frac12}} \;$ with respect to \(z\).
  2. The {\em signed curvature} \(\kappa\) of the curve \(y=\f(x)\) is defined by \[ \kappa = \frac {\f''(x)}{\big({1+ (\f'(x))^2\big)^{\frac32}}} \,.\] Use this definition to determine all curves for which the signed curvature is a non-zero constant. For these curves, what is the geometrical significance of \(\kappa\)?

Show Solution
  1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
  2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.
2016 Paper 2 Q6
D: 1600.0 B: 1484.0

This question concerns solutions of the differential equation \[ (1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\, \tag{\(*\)} \] where \(k\) is a positive integer. For each value of \(k\), let \(y_k(x)\) be the solution of \((*)\) that satisfies \(y_k(1)=1\); you may assume that there is only one such solution for each value of \(k\).

  1. Write down the differential equation satisfied by \(y_1(x)\) and verify that \(y_1(x) = x\,\).
  2. Write down the differential equation satisfied by \(y_2(x)\) and verify that \(y_2(x) = 2x^2-1\,\).
  3. Let \(z(x) = 2\big(y_n(x)\big)^2 -1\). Show that \[ (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\, \] and hence obtain an expression for \(y_{2n}(x)\) in terms of \(y_n(x)\).
  4. Let \(v(x) = y_n\big(y_m(x)\big)\,\). Show that \(v(x) = y_{mn}(x)\,\).

Show Solution
  1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
  2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
  3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
  4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.
[Note: this is another question about Chebyshev polynomials, and we have proven that we can compose them nicely. This might be more easily proven as \(T_n(x) = \cos(n \cos^{-1} x)\) and so \(T_n(T_m(x)) = \cos (n \cos^{-1}( \cos (m \cos^{-1} x))) = \cos(nm \cos^{-1}x) = T_{nm}(x)\)]
2013 Paper 1 Q7
D: 1516.0 B: 1516.0

  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 \(x=1\).

Show 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*}
2011 Paper 1 Q7
D: 1500.0 B: 1500.0

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 \(\alpha\).

Show 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.
2010 Paper 3 Q8
D: 1700.0 B: 1531.5

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 \[ (1+\cos x +2 \sin x) \frac {\d y}{\d x} +(\sin x -2 \cos x)y = 5 - 3 \cos x + 4 \sin x\,. \]

Show 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*}
2008 Paper 2 Q7
D: 1600.0 B: 1472.0

  1. By writing \(y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}\), where \(u\) is a function of \(x\), find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = xy + \frac x {1+x^2} \] for which \(y=1\) when \(x=0\).
  2. Find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^2y + \frac {x^2 } {1+x^3} \] for which \(y=1\) when \(x=0\).
  3. Give, without proof, a conjecture for the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n} \] for which \(y=1\) when \(x=0\), where \(n\) is an integer greater than 1.

2007 Paper 2 Q6
D: 1600.0 B: 1469.4

  1. Differentiate \(\ln\big (x+\sqrt{3+x^2}\,\big)\) and \(x\sqrt{3+x^2}\) and simplify your answers. Hence find \(\int \! \sqrt{3+x^2}\, \d x\).
  2. Find the two solutions of the differential equation \[ 3\left(\frac{\d y}{\d x}\right)^{\!2} + 2 x \frac {\d y}{\d x} =1 \] that satisfy \(y=0\) when \(x=1\).

Show Solution
  1. \(\,\) \begin{align*} && y &= \ln (x + \sqrt{3+x^2}) \\ \Rightarrow && y' &= \frac{1}{x + \sqrt{3+x^2}} \cdot \left (1 + \frac{x}{\sqrt{3+x^2}} \right) \\ &&&= \frac{1}{\sqrt{3+x^2}} \\ \\ && y &= x\sqrt{3+x^2} \\ && y' &= \sqrt{3+x^2} + \frac{x^2}{\sqrt{3+x^2}} \\ &&&= 2\sqrt{3+x^2} - \frac{3}{\sqrt{3+x^2}} \\ \\ \Rightarrow && \sqrt{3+x^2} &= \frac12(x \sqrt{3+x^2})' + \frac32(\ln(x+\sqrt{3+x^2})' \\ \Rightarrow && \int \sqrt{3+x^2}\, \d x &= \frac12x\sqrt{3+x^2} + \frac32 \ln (x+\sqrt{3+x^2}) + C \end{align*}
  2. \(\,\) \begin{align*} && 3 \left ( \frac{\d y}{\d x} \right)^2 + 2x \frac{\d y}{\d x} &= 1 \\ && \frac{\d y}{\d x} &= \frac{-x \pm \sqrt{x^2+3} }3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac16x\sqrt{3+x^2} \pm \frac12 \ln (x+\sqrt{3+x^2}) + C \\ y = 0, x = 1: && 0 &= -\frac16 \pm \frac13 \pm \frac12 \ln 3 \\ \Rightarrow && y &= -\frac{x^2}{6} \pm \frac12x\sqrt{3+x^2} \pm \frac32 \ln (x+\sqrt{3+x^2}) + \frac16 \mp \frac13 \mp \frac12 \ln 3 \end{align*}
2005 Paper 1 Q8
D: 1500.0 B: 1484.0

Show that, if \(y^2 = x^k \f(x)\), then $\displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = ky^2 + x^{k+1} \frac{\mathrm{d}\f }{ \mathrm{d}x}$\,.

  1. By setting \(k=1\) in this result, find the solution of the differential equation \[ \displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = y^2 + x^2 - 1 \] for which \(y=2\) when \(x=1\). Describe geometrically this solution.
  2. Find the solution of the differential equation \[ 2x^2y\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \ln(x) - xy^2 \] for which \(y=1\) when \(x=1\,\).

2005 Paper 2 Q8
D: 1600.0 B: 1516.0

For \(x \ge 0\) the curve \(C\) is defined by $$ {\frac{\d y} {\d x}} = \frac{x^3y^2}{(1 + x^2)^{5/2}} $$ with \(y = 1\) when \(x=0\,\). Show that \[ \frac 1 y = \frac {2+3x^2}{3(1+x^2)^{3/2}} +\frac13 \] and hence that for large positive \(x\) $$ y \approx 3 - \frac 9 x\;. $$ Draw a sketch of \(C\). On a separate diagram, draw a sketch of the two curves defined for \(x \ge 0\) by $$ \frac {\d z} {\d x} = \frac{x^3z^3}{2(1 + x^2)^{5/2}} $$ with \(z = 1\) at \(x=0\) on one curve, and \(z = -1\) at \(x=0\) on the other.

Show Solution
\begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}
TikZ diagram
\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:
TikZ diagram
2005 Paper 3 Q2
D: 1700.0 B: 1502.0

Find the general solution of the differential equation \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;\), where \(a\ne0\,\), and show that it can be written in the form \(\displaystyle y^2(x^2+a^2)= c^2\,\), where \(c\) is an arbitrary constant. Sketch this curve. Find an expression for \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)\) and show that \[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;. \]

  1. Show that, if \(0 < c < a^2\), the points on the curve whose distance from the origin is least are \(\displaystyle \l 0\,,\;\pm \frac{c}{a}\r\).
  2. If \(c > a^2\), determine the points on the curve whose distance from the origin is least.

Show Solution
\begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).
TikZ diagram
\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}
  1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
  2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)
TikZ diagram
2004 Paper 2 Q5
D: 1600.0 B: 1516.0

Evaluate \(\int_0^{{\pi}} x \sin x\,\d x\) and \(\int_0^{{\pi}} x \cos x\,\d x\;\). The function \(\f\) satisfies the equation \begin{equation} \f(t)=t + \int_0^{{\pi}} \f(x)\sin(x+t)\,\d x\;. \tag{\(*\)} \end{equation} Show that \[ \f(t)=t + A\sin t + B\cos t\;, \] where \(A= \int_0^{{\pi}}\,\f(x)\cos x\,\d x\;\) and \(B= \int_0^{{\pi}}\,\f(x)\sin x\,\d x\;\). Find \(A\) and \(B\) by substituting for \(\f(t)\) and \(\f(x)\) in \((*)\) and equating coefficients of \(\sin t\) and \(\cos t\,\).

Show Solution
\begin{align*} && I &= \int_0^\pi x \sin x \d x \\ &&&= \left [ -x \cos x \right]_0^\pi + \int_0^{\pi} \cos x \d x \\ &&&= \pi \\ \\ && J &= \int_0^\pi x \cos x \d x \\ &&&= \left [ x \sin x \right]_0^\pi - \int_0^\pi \sin x \d x \\ &&&= -2 \end{align*} \begin{align*} && f(t) &= t + \int_0^\pi f(x) \sin (x+t) \d x \\ &&&= t + \int_0^\pi f(x) \left ( \sin t \cos x + \cos t \sin x \right) \d x \\ &&&= t + \sin t \int_0^{\pi} f(x) \cos x \d x + \cos t \int_0^{\pi} f(x) \sin x \d x \\ \\ && A &= \int_0^\pi (x + A \sin x + B \cos x) \cos x \d x \\ &&&= -2+ \frac{\pi}{2} B \\ && B &= \int_0^{\pi} (x + A \sin x + B \cos x ) \sin x \d x \\ &&&= \pi + \frac{\pi}{2} A \\ \Rightarrow && (A,B) &= (-2,0) \end{align*}
2004 Paper 2 Q8
D: 1600.0 B: 1483.3

Let \(x\) satisfy the differential equation $$ \frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n} $$ and the condition \(x=0\) when \(t=0 \,\).

  1. Solve the equation in the case \(n=1\) and sketch the graph of the solution for \(t > 0 \,\).
  2. Prove that \(1-x < (1-x^2)^{1/2} \) for \(0 < x < 1 \,\). Use this result to sketch the graph of the solution in the case \(n=2\) for \(0 < t < \frac12 \pi \,\), using the same axes as your previous sketch. By setting \(x=\sin y\,\), solve the equation in this case.
  3. Use the result (which you need not prove) \[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \] to sketch, without solving the equation, the graph of the solution of the equation in the case \(n=3\) using the same axes as your previous sketches. Use your sketch to show that \(x=1\) at a value of \(t\) less than \(\frac12 \pi \,\).

Show Solution
  1. \(\,\) \begin{align*} && \dot{x} &= (1-x) \\ \Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\ \Rightarrow && -\ln |1-x| &= t + C \\ t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\ \Rightarrow && -\ln|1-x| &= t \\ \Rightarrow && 1-x&= e^{-t} \\ \Rightarrow && x &= 1-e^{-t} \end{align*}
    TikZ diagram
  2. Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
    TikZ diagram
    \begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
  3. \(\,\)
    TikZ diagram
    We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)
2003 Paper 1 Q8
D: 1516.0 B: 1484.0

A liquid of fixed volume \(V\) is made up of two chemicals \(A\) and \(B\,\). A reaction takes place in which \(A\) converts to \(B\,\). The volume of \(A\) at time \(t\) is \(xV\) and the volume of \(B\) at time \(t\) is \(yV\) where \(x\) and \(y\) depend on \(t\) and \(x+y=1\,\). The rate at which \(A\) converts into \(B\) is given by \(kVxy\,\), where \(k\) is a positive constant. Show that if both \(x\) and \(y\) are strictly positive at the start, then at time \(t\) \[ y= \frac {D\e^{kt}}{1+D \e^{kt}} \;, \] where \(D\) is a constant. Does \(A\) ever completely convert to \(B\,\)? Justify your answer.

Show Solution
We have \(\dot{A} = -kVxy\) or \(\dot{x}V = -kVxy\), ie \(\dot{x} = -kxy\) and similarly \(\dot{y} = kxy = k(1-y)y\). \begin{align*} && \frac{\d y}{\d t} &= ky(1-y) \\ \Rightarrow && \int k \d t &= \int \frac{1}{y(1-y)} \d y \\ \Rightarrow && kt &= \int \left ( \frac{1}{y} + \frac{1}{1-y} \right) \d y \\ &&&= \ln y - \ln (1-y) + C\\ \Rightarrow && kt &= \ln \frac{y}{D(1-y)} \\ \Rightarrow && De^{kt} &= \frac{y}{1-y} \\ \Rightarrow && y(1+De^{kt}) &= De^{kt} \\ \Rightarrow && y &= \frac{De^{kt}}{1+De^{kt}} \end{align*} As \(t \to \infty\) \(y \to \frac{D}{D} = 1\) so depending on how fine grained we want to go we might say that 'yes it completely converts' when there is an immeasurably small amount of \(A\) left, or we might say it doesn't since it only tends to \(1\) and never actually reaches it.
2003 Paper 2 Q8
D: 1600.0 B: 1516.0

It is given that \(y\) satisfies $$ {{\d y} \over { \d t}} + k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;, $$ where \(k\) is a constant, and \(y=A \) when \(t=0\,\), where \(A\) is a positive constant. Find \(y\) in terms of \(t\,\), \(k\) and \(A\,\). Show that \(y\) has two stationary values whose ratio is \((3/2)^{6k}\e^{-5{k}/2}.\) Describe the behaviour of \(y\) as \(t \to +\infty\) for the case where \(k> 0\) and for the case where \(k<0\,.\) In separate diagrams, sketch the graph of \(y\) for \(t>0\) for each of these cases.

Show Solution
\begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.
TikZ diagram
2002 Paper 2 Q8
D: 1600.0 B: 1500.0

Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

2002 Paper 3 Q6
D: 1700.0 B: 1484.9

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

  1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
  2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).
Show also that \(y = 1\) and \(y = -1\) are solutions of the differential equation. Sketch all these solution curves on a single set of axes.

Show Solution
\begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}
  1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
  2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)
Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.
TikZ diagram
2001 Paper 1 Q7
D: 1500.0 B: 1516.0

In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

  1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
  2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
  3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).

Show Solution
  1. \(\,\)
    TikZ diagram
    Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
  2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
  3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.
2001 Paper 2 Q5
D: 1600.0 B: 1485.5

The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)

2001 Paper 2 Q8
D: 1600.0 B: 1488.2

The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).

  1. Give an example of such a function.
  2. The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
  3. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)

Show Solution
  1. \(f(x) = \lfloor x \rfloor+1\)
  2. Clearly \(\displaystyle F(x) = \int_0^x f(t) \d t\), in particular: \begin{align*} && F(n) &= \int_0^n f(t) \d t \\ &&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\ &&&= n \int_{0}^1 f(t) \d t\\ &&&= n F(1) \end{align*}
  3. \(\,\) \begin{align*} && 0 &= \frac{\d y}{\d x} +f(x) y \\ \Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\ \Rightarrow && -F(x) & = \ln y + C \\ x=0,y=1: && C &= -F(0) \\ \Rightarrow && y &= \exp(F(0)-F(x)) \end{align*} Well this \(F(0)-F(x)\) is equivalent to \(-F(x)\) where \(F(0) = 0\), in particular \(F(n) = nF(1)\), so \(y(n) = e^{-nF(1)}\) which tends to zero as long as \(F(1) > 0\), but since \(f(x) > 0\) for all \(x\) this must be true.
2001 Paper 3 Q7
D: 1700.0 B: 1516.0

Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

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\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}
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1998 Paper 1 Q8
D: 1516.0 B: 1500.0

Fluid flows steadily under a constant pressure gradient along a straight tube of circular cross-section of radius \(a\). The velocity \(v\) of a particle of the fluid is parallel to the axis of the tube and depends only on the distance \(r\) from the axis. The equation satisfied by \(v\) is \[\frac{1}{r}\frac{{\mathrm d}\ }{{\mathrm d}r} \left(r\frac{{\mathrm d}v}{{\mathrm d}r}\right) =-k,\] where \(k\) is constant. Find the general solution for \(v\). Show that \(|v|\rightarrow\infty\) as \(r\rightarrow 0\) unless one of the constants in your solution is chosen to be~\(0\). Suppose that this constant is, in fact, \(0\) and that \(v=0\) when \(r=a\). Find \(v\) in terms of \(k\), \(a\) and \(r\). The volume \(F\) flowing through the tube per unit time is given by \[F=2\pi\int_{0}^{a}rv\,{\mathrm d}r. \] Find \(F\).