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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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2003 Paper 3 Q10
D: 1700.0 B: 1500.0
differential equation variable force acceleration proportional to position and velocity separation of variables Newton's second law asymptotic behaviour trigonometric nonlinear ODE

A particle moves along the \(x\)-axis in such a way that its acceleration is \(kx \dot{x}\,\) where \(k\) is a positive constant. When \(t = 0\), \(x = d\) (where \(d>0\)) and \(\dot{x} =U\,\).

  1. Find \(x\) as a function of \(t\) in the case \(U = kd^2\) and show that \(x\) tends to infinity as \(t\) tends to \(\displaystyle \frac{\pi }{2 dk}\,\).
  2. If \(U < 0\), find \(x\) as a function of \(t\) and show that it tends to a limit, which you should state in terms of \(d\) and \(U\,\), as \(t\) tends to infinity.

Solution:

  1. \(\,\) \begin{align*} && \ddot{x} &= kx \dot{x} \\ \Rightarrow && \frac{\d v}{\d x} \dot{x} &= k x \dot{x} \\ \Rightarrow && \int \d v &= \int k x \d x \\ \Rightarrow && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = kd^2: && kd^2 &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2+d^2) \\ \Rightarrow && \frac{\d x}{\d t} &= \frac12k(x^2+d^2) \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2+d^2)} \d x \\ &&&= \frac{2}{kd}\tan^{-1} \frac{x}{d} \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d} + C' \\ t = 0, x = d: && 0 &= \frac{\pi}{2kd} + C' \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d}-\frac{\pi}{2kd} \end{align*} As \(x \to \infty\), \(t \to \frac{2}{kd} \frac{\pi}{2} - \frac{\pi}{2kd} = \frac{\pi}{2kd} \)
  2. \(\,\) \begin{align*} && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = U && U &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2-d^2)+U \\ \Rightarrow && \frac{\d x}{\d t} &=\frac12k(x^2-d^2)+U \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2-d^2)+U} \d x \\ && &=\frac{2}{k} \int \frac{1}{x^2-d^2+\frac{2U}k} \d x \\ &&&= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} + C'' \\ t = 0, \dot{x} = d: && 0 &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-d}{d+\sqrt{d^2-\frac{2U}k}} + C'' \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \left ( \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \frac{d+\sqrt{d^2-\frac{2U}k}}{ \sqrt{d^2-\frac{2U}k}-d} \right ) \end{align*} as \(t \to \infty\) the denominator needs to head to \(0\), ie \(x \to -\sqrt{d^2-\frac{2U}k}\)

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1990 Paper 2 Q8
D: 1600.0 B: 1500.0
integration integration by parts differential equation second order boundary conditions nonlinear ODE Fundamental Theorem of Calculus periodic boundary conditions

The functions \(\mathrm{x}\) and \(\mathrm{y}\) are related by \[ \mathrm{x}(t)=\int_{0}^{t}\mathrm{y}(u)\,\mathrm{d}u, \] so that \(\mathrm{x}'(t)=\mathrm{y}(t)\). Show that \[ \int_{0}^{1}\mathrm{x}(t)\mathrm{y}(t)\,\mathrm{d}t=\tfrac{1}{2}\left[\mathrm{x}(1)\right]^{2}. \] In addition, it is given that \(\mbox{y}(t)\) satisfies \[ \mathrm{y}''+(\mathrm{y}^{2}-1)\mathrm{y}'+\mathrm{y}=0,\mbox{ }(*) \] with \(\mathrm{y}(0)=\mathrm{y}(1)\) and \(\mathrm{y}'(0)=\mathrm{y}'(1)\). By integrating \((*)\), prove that \(\mathrm{x}(1)=0.\) By multiplying \((*)\) by \(\mathrm{x}(t)\) and integrating by parts, prove the relation \[ \int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t=\tfrac{1}{3}\int_{0}^{1}\left[\mathrm{y}(t)\right]^{4}\,\mathrm{d}t. \] Prove also the relation \[ \int_{0}^{1}\left[\mathrm{y}'(t)\right]^{2}\,\mathrm{d}t=\int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t. \]

Solution: Consider \(\frac12 x(t)^2\) then differentiating we obtain \(x(t)x'(t) = x(t)y(t)\). Also note that \(x(0) = \int_0^0 y(u) \d u = 0\) Therefore, \begin{align*} \int_0^1 x(t)y(t) \d t &= \left [ \frac12 x(t)^2 \right]_0^1 \\ &= \frac12[x(1)]^2 \end{align*} \begin{align*} && 0 &= y'' + (y^2-1)y' + y \\ \Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\ &&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\ &&&= x(1) \end{align*} Therefore \(x(1) = 0\). \begin{align*} && 0 &= xy'' + (y^2-1)y' x+ yx \\ \Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\ &&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\ &&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\ \Rightarrow && \int_0^1 [y(t)]^2 \d t &= \frac13 \int_0^1 [y(t)]^4 \d t \end{align*} \begin{align*} && 0 &= yy'' + (y^2-1)y' y+ y^2 \\ \Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\ &&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ &&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ \Rightarrow && \int_0^1 [y'(t)]^2 \d t &= \int_0^1 [y(t)]^2 \d t \end{align*}

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