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2014 Paper 2 Q5
D: 1600.0 B: 1486.1
differential equation first order homogeneous ODE substitution y=xu translation of axes quadratic in x and y initial value problem

Given that \(y=xu\), where \(u\) is a function of \(x\), write down an expression for \(\dfrac {\d y}{\d x}\).

  1. Use the substitution \(y=xu\) to solve \[ \frac {\d y}{\d x} = \frac {2y+x}{y-2x} \] given that the solution curve passes through the point \((1,1)\). Give your answer in the form of a quadratic in \(x\) and \(y\).
  2. Using the substitutions \(x=X+a\) and \(y=Y+b\) for appropriate values of \(a\) and \(b\), or otherwise, solve \[ \frac {\d y}{\d x} = \frac {x-2y-4} {2x+y-3}\,, \] given that the solution curve passes through the point \((1,1)\).

Solution: \begin{align*} \frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\ &= \frac{\d }{\d x} \l xu \r \\ &\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\ &= u + x \frac{\d u}{\d x} \end{align*} \begin{questionparts} \item \begin{align*} && \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\ && u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\ && x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\ \Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\ && \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\ && -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\ (x,y) = (1,1): && - \ln 2 &= C \\ \Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\ \Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\ \Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\ &&&= 5x^2 - (y-2x)^2 \\ &&&= x^2+4xy-y^2 \end{align*} \item It would be convienient if \(x-2y -4 = X-2Y\) and \(2x+y-3 = 2X+Y\), ie \(a-2b = 4\) and \(2a+b = 3\), ie \(a = 2, b = -1\). Now our differential equation is: \begin{align*} && \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\ && \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y} \end{align*} This is the same differential equation we have already solved, just with the roles of \(x\) and \(y\) interchanged with \(Y\) and \(X\) and with the point \((0,3)\) being on the curve, ie: \(Y^2 + 4XY-X^2 = c\) and \(c = 9\), therefore our equation is: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]

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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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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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2003 Paper 3 Q8
D: 1700.0 B: 1516.0
first order differential equation implicit differentiation exact equation integrating factor homogeneous ODE curve gradient

  1. Show that the gradient at a point \(\l x\,, \, y \r\) on the curve \[ \l y + 2x \r^3 \l y - 4x \r = c\;, \] where \(c\) is a constant, is given by \[ \frac{\d y}{\d x} = \frac{16 x -y}{2y-5x} \;. \]
  2. By considering the derivative with respect to \(x\) of \(\l y + ax \r^n \l y + bx \r\,\), or otherwise, find the general solution of the differential equation \[ \frac{\mathrm{d}y}{ \mathrm{d}x} = \frac{10x - 4y}{ 3x - y}\;. \]

Solution:

  1. \begin{align*} && c &= \left (y+2x \right)^3\left (y-4x \right) \\ \Rightarrow && 0 &= 3\left (y+2x \right)^2\left (y-4x \right)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right)^3 \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow && 0 &= 3(y-4x)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right) \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow &&&= \frac{\d y}{\d x} \left (3(y-4x) + (y+2x) \right) + 6(y-4x)-4(y+2x) \\ &&&= \frac{\d y}{\d x} \left ( 4y-10x\right) + 2y-32x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{16x-y}{2y-5x} \end{align*}
  2. \begin{align*} && c &= \left ( y + ax \right)^n \left ( y + bx \right) \\ \Rightarrow && 0 &= n\left ( y + ax \right)^{n-1} \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right)^{n} \left ( \frac{\d y}{ \d x}+b \right) \\ \Rightarrow && 0 &= n \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right) \left ( \frac{\d y}{ \d x}+b \right) \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + an(y+bx) + by+bax \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + (an+b)y+ab(n+1)x \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{(an+b)y+ab(n+1)x}{(n+1)y+(nb+a)x} \end{align*} We must have \(ab = 10, a+b = -7\) so say \(a=-5,b=-2,n=2\) and we have \((y-5x)^2(y-2) = c\) is our general solution to the differential equation

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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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