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2025 Paper 2 Q7
D: 1500.0 B: 1500.0
differential equation second order nonlinear ODE reduction of order separation of variables particle motion initial value problem long-term behaviour

The differential equation \[\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}\] describes the motion of a particle with position \(x(t)\) at time \(t\). At \(t = 0\), \(x = a\), where \(a > 0\).

  1. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2\) when \(t = 0\). What happens to the particle as \(t\) increases from 0?
  2. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 + p\) when \(t = 0\), where \(p > 0\). What happens to the particle as \(t\) increases from 0?
  3. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 - q^2\) when \(t = 0\), where \(q > 0\). What happens to the particle as \(t\) increases from 0? Give conditions on \(a\) and \(q\) for the different cases which arise.

Solution: Let \(v = \frac{\d x}{\d t}\) and notice that \(\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}\). Also notice that: \begin{align*} && v \frac{\d v}{\d x} &= 2x v \\ \Rightarrow && \frac{\d v}{\d x} &= 2x \\ \Rightarrow && v &= x^2 + C \\ \Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\ \end{align*}

  1. When \(t = 0, \frac{\d x}{\d t} = a^2\) so \(C = 0\), therefore \(\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k\) and so \(k = a^{-1}\) and \(x = \frac{a}{1-at}\). As \(t\) increases from \(0\) the particle heads to infinity at an increasing rate, `reaching' infinity around \(t=\frac{1}{a}\)
  2. When \(t = 0, \frac{\d x}{\d t} = a^2 + p\) so \(C = p\). Therefore \(\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c\). When \(c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)\), so \begin{align*} && t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\ &&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\ \Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\ \Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\ \Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\ \Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)} \end{align*} The particle heads to \(\frac{\sqrt{p}}{a}\).
  3. When \(t = 0, \frac{\d x}{\d t} = a^2-q^2\) so \(C = -q^2\). Therefore \begin{align*} && \frac{\d x}{\d t} &= x^2 -q^2 \\ \Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\ &&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\ &&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\ &&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\ \Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\ \underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\ \Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\ \Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\ \Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}} \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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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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1993 Paper 2 Q3
D: 1600.0 B: 1500.0
differential equation Bernoulli equation substitution Riccati equation first order ODE second order linear ODE reduction of order

  1. Solve the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}-y-3y^{2}=-2 \] by making the substitution \(y=-\dfrac{1}{3u}\dfrac{\mathrm{d}u}{\mathrm{d}x}.\)
  2. Solve the differential equation \[ x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}+xy+x^{2}y^{2}=1 \] by making the substitution \[ y=\frac{1}{x}+\frac{1}{v}, \] where \(v\) is a function of \(x\).

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