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2007 Paper 3 Q8
D: 1700.0 B: 1487.5
differential equation second order variable coefficients Riccati equation substitution particular solution general solution first order nonlinear ODE

  1. Find functions \({\rm a}(x)\) and \({\rm b}(x)\) such that \(u=x\) and \(u=\e^{-x}\) both satisfy the equation $$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$ For these functions \({\rm a}(x)\) and \({\rm b}(x)\), write down the general solution of the equation. Show that the substitution \(y= \dfrac 1 {3u} \dfrac {\d u}{\d x}\) transforms the equation \[ \frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)} \tag{\(*\)} \] into \[ \frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x} u=0 \] and hence show that the solution of equation (\(*\)) that satisfies \(y=0\) at \(x=0\) is given by \(y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}\).
  2. Find the solution of the equation $$ \frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x} $$ that satisfies \(y=2\) at \(x=0\).

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2001 Paper 1 Q8
D: 1500.0 B: 1487.5
differential equation second order homogeneous ODE simultaneous equations trigonometric superposition principle variable coefficients

Given that \(y=x\) and \(y=1-x^2\) satisfy the differential equation $$ \frac{\d^2 {y}}{\d x^2} + \p(x) \frac{\d {y}}{\d x} + \q(x) {y}=0\;, \tag{*} $$ show that \(\p(x)= -2x(1+x^2)^{-1}\) and \(\q(x) = 2(1+x^2)^{-1}\). Show also that \(ax+b(1-x^2)\) satisfies the differential equation for any constants \(a\) and \(b\). Given instead that \(y=\cos^2(\frac{1}{2}x^2)\) and \(y=\sin^2(\frac{1}{2}x^2)\) satisfy the equation \((*)\), find \(\p(x)\) and \(\q(x)\).

Solution: \begin{align*} && y &= x \\ && y' &= 1 \\ && y'' &= 0 \\ \Rightarrow && 0 &= 0 + p(x) + xq(x) \tag{1} \\ \\ && y &= 1-x^2 \\ && y' &= -2x \\ && y'' &= -2 \\ \Rightarrow && 0 &= -2 -2x p(x)+(1-x^2)q(x) \tag{2}\\ \\ 2x*(1) +(2): && 2 &= (2x^2+1-x^2) q(x) \\ \Rightarrow && q(x) &= 2(1+x^2)^{-1} \\ \Rightarrow && p(x) &= -2x(1+x^2)^{-1} \tag{by (1)} \end{align*} \begin{align*} && \frac{\d^2}{\d x^2} \left (a x + b(1-x^2) \right) + p(x) \frac{\d}{\d x} \left (a x + b(1-x^2) \right)+q(x) \left (a x + b(1-x^2) \right) \\ &&= a \frac{\d^2 x}{\d x^2} + b \frac{\d^2}{\d x^2} \left ( 1- x^2 \right) + ap(x) \frac{\d x}{ \d x} + bp(x) \frac{\d }{\d x} \left ( 1- x^2 \right) + aq(x) x + bq(x)(1-x^2) \\ &&= a \left (\frac{\d^2 x}{\d x^2}+ p(x) \frac{\d x}{ \d x} +q(x)x\right)+b \left ( \frac{\d^2}{\d x^2} \left ( 1- x^2 \right)+ p(x) \frac{\d }{\d x} \left ( 1- x^2 \right)+q(x)(1-x^2)\right) &= 0 \end{align*} \begin{align*} && y &= \cos^2(\tfrac12 x^2) = \frac12 \left (1 + \cos(x^2) \right) \\ && y' &= -x \sin(x^2) \\ && y'' &= -2x^2 \cos(x^2)-\sin(x^2) \\ \Rightarrow && 0 &= -2x^2 \cos(x^2)-\sin(x^2)+p(x)(-x \sin(x^2)) +\frac12 \left (1 + \cos(x^2) \right)q(x) \\ \Rightarrow && 2x^2\cos(x^2)+\sin(x^2) &= -x \sin(x^2) p(x) + \frac12(1 + \cos(x^2)) q(x) \tag{3}\\ \\ && y &= \sin^2(\tfrac12 x^2) = \frac12 \left ( 1 - \cos (x^2) \right) \\ && y' &= x\sin(x^2) \\ && y'' &= 2x^2 \cos(x^2)+\sin(x^2) \\ \Rightarrow && 0 &= 2x^2 \cos(x^2)+\sin(x^2) +p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x)\\ \Rightarrow && -2x^2 \cos(x^2)-\sin(x^2) &= p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x) \tag{4}\\ (3)+(4): && 0 &= q(x) \\ \Rightarrow && p(x) &= -\frac{2x^2 \cos(x^2)+\sin(x^2)}{x \sin(x^2)} \end{align*}

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1995 Paper 1 Q8
D: 1500.0 B: 1532.8
second order differential equation variable coefficients particular integral complementary function linear combination simultaneous equations superposition principle

Find functions \(\mathrm{f,g}\) and \(\mathrm{h}\) such that \[ \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+\mathrm{f}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{g}(x)y=\mathrm{h}(x)\tag{\ensuremath{*}} \] is satisfied by all three of the solutions \(y=x,y=1\) and \(y=x^{-1}\) for \(0 < x < 1.\) If \(\mathrm{f,g}\) and \(\mathrm{h}\) are the functions you have found in the first paragraph, what condition must the real numbers \(a,b\) and \(c\) satisfy in order that \[ y=ax+b+\frac{c}{x} \] should be a solution of \((*)\)?

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