First order differential equations (integrating factor)

Showing 1-12 of 12 problems
2018 Paper 2 Q8
D: 1600.0 B: 1484.0

  1. Use the substitution \(v= \sqrt y\) to solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac12} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_1(x)\) that satisfies \(y_1(0)=0\,\).
  2. Solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac23} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_2(x)\) that satisfies \(y_2(0)=0\,\).
  3. In the case \(\alpha=\beta\), sketch \(y_1(x)\) and \(y_2(x)\) on the same axes, indicating clearly which is \(y_1(x)\) and which is \(y_2(x)\). You should explain how you determined the positions of the curves relative to each other.

Show Solution
  1. Suppose \(v = \sqrt{y} \Rightarrow v^2 = y \Rightarrow 2v v' = y'\) so \begin{align*} && 2vv' &= \alpha v-\beta v^2 \\ \Rightarrow && 2v' &= \alpha - \beta v \\ \Rightarrow && v' + \frac{\beta}{2} v &= \frac{\alpha}{2} \\ \Rightarrow && \frac{\d}{\d t} \left (e^{\beta t/2} v \right) &= \frac{\alpha}{2} e^{\beta t/2} \\ \Rightarrow && e^{\beta t/2} v &=C+ \frac{\alpha}{\beta}e^{\beta t /2} \\ \Rightarrow && v &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ y(0) = 0: && 0 &= C+\frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= \frac{\alpha}{\beta} \left (1-e^{-\beta t/2} \right) \\ \Rightarrow && y &= \frac{ \alpha^2}{\beta^2} \left (1-e^{-\beta t/2} \right)^2 \end{align*}
  2. Try \(v = y^{1/3} \Rightarrow v^3 = y \Rightarrow 3v^2 v' = y'\) so \begin{align*} && y' &= \alpha v^2 - \beta y \\ \Rightarrow && 3v^2v' &= \alpha v^2 - \beta v^3 \\ \Rightarrow && v' +\frac{\beta}{3} v &= \frac{\alpha}{3} \\ \Rightarrow && (v e^{\beta t/3})' &= \frac{\alpha}{3}e^{\beta t/3} \\ \Rightarrow && v &= Ce^{-\beta t/3} + \frac{\alpha}{\beta} \\ v(0) = 0: && v &= \frac{\alpha}{\beta} \left (1 - e^{-\beta t/3} \right) \\ \Rightarrow && y &= \frac{\alpha^3}{\beta^3} \left (1 - e^{-\beta t/3} \right) ^3 \end{align*}
  3. \(y_1 = (1-e^{-\beta t/2})^2, y_2 = (1-e^{-\beta t/3})^3\)
    TikZ diagram
    By considering the differential equation, notice that \(0 < y_i < 1\) so \(y^{1/2} > y^{2/3}\) and therefore \(y_1' > y_2'\) and so \(y_1\) increases faster.
2018 Paper 2 Q10
D: 1600.0 B: 1500.0

A uniform elastic string lies on a smooth horizontal table. One end of the string is attached to a fixed peg, and the other end is pulled at constant speed \(u\). At time \(t=0\), the string is taut and its length is \(a\). Obtain an expression for the speed, at time \(t\), of the point on the string which is a distance \(x\) from the peg at time \(t\). An ant walks along the string starting at \(t=0\) at the peg. The ant walks at constant speed \(v\) along the string (so that its speed relative to the peg is the sum of \(v\) and the speed of the point on the string beneath the ant). At time \(t\), the ant is a distance \(x\) from the peg. Write down a first order differential equation for \(x\), and verify that \[ \frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,. \] Show that the time \(T\) taken for the ant to reach the end of the string is given by \[uT = a(\e^k-1)\,,\] where \(k=u/v\). On reaching the end of the string, the ant turns round and walks back to the peg. Find in terms of \(T\) and \(k\) the time taken for the journey back.

Show Solution
Points always maintain a constant fraction of the distance from the start, so the point distance \(x\) from the start at time \(t\) is moving with speed \(\frac{x}{a+ut} u\) The point is moving with speed \(v+\frac{x}{a+ut} u\) or in other words \begin{align*} && \frac{\d x}{\d t} &= v + \frac{x}{a+ut}u \\ \Rightarrow && \frac{\d x }{\d t} - \frac{u}{a+ut} x &= v \\ \Rightarrow && \frac{1}{a+ut} \frac{\d x}{\d t} - \frac{u}{(a+ut)^2} x &= \frac{1}{a+ut} v\\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{x}{a+ut} \right) &= \frac{v}{a+ut} \\ \Rightarrow && \frac{x}{a+ut} &=\frac{v}{u} \ln (a + ut) + C \\ t = 0, x = 0: && 0 &= \frac{v}{u} \ln a + C \\ \Rightarrow && x &= (a+ut) \frac{v}{u} \ln \left ( \frac{a+ut}{a} \right) \\ \\ \Rightarrow && 1 &= \frac{v}{u} \ln \left ( \frac{a+uT}{a} \right) \\ \Rightarrow && e^k &= 1 + \frac{uT}{a} \\ \Rightarrow && uT &= a(e^k-1) \end{align*} On the return journey, we have \begin{align*} && \frac{\d x}{\d t} &= \frac{x}{a+ut}u - v \\ \Rightarrow && \frac{\d x}{\d t} - \frac{u}{a+ut} x &= - v \\ \Rightarrow && \frac{\d }{\d x} \left ( \frac{x}{a+ut} \right) &= -\frac{v}{a+ut} \\ \Rightarrow &&f &= -\frac{v}{u} \ln(a+ut) + K \\ t = T, f = 1: && 1 &= -\frac{v}{u}\ln(a+uT) + K \\ \Rightarrow && f &= 1+\frac{v}{u}\ln \left ( \frac{a+uT}{a+ut} \right) \\ \Rightarrow && 0 &= 1+\frac{v}{u} \ln \left ( \frac{a+uT}{a+uT_2} \right) \\ \Rightarrow && e^k &= \frac{a+uT_2}{a+uT}\\ \Rightarrow && uT_2 &= (a+uT)e^k - a \\ \Rightarrow && T_2 - T &= \frac{1}{u} \left ( (a+uT)e^k - a - uT\right) \\ &&&= \frac{1}{u} \left ((a+a(e^k-1))e^k-a-a(e^k-1) \right) \\ &&&= \frac{1}{u} \left (ae^{2k} -ae^k \right) \\ &&&= \frac{ae^k}{u} \left ( e^k-1 \right) \\ &&&= Te^k \end{align*}
2014 Paper 2 Q5
D: 1600.0 B: 1486.1

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 \((1,1)\).

Show 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: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]
2012 Paper 1 Q8
D: 1516.0 B: 1484.0

  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 \[ y \frac{\d y}{\d x} +6x +5y=0\, \quad (x\ne0). \]

Show 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*}
2004 Paper 3 Q8
D: 1700.0 B: 1484.0

Show that if \[ {\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y} \] then the substitution \(u = y^2\) gives a linear differential equation for \(u(x)\,\). Hence or otherwise solve the differential equation \[ {\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;. \] Determine the solution curves of this equation which pass through \((1 \,, 1)\,\), \((2\, , 2)\) and \((4 \, , 4)\) and sketch graphs of all three curves on the same axes.

Show Solution
\begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)
TikZ diagram
2003 Paper 3 Q8
D: 1700.0 B: 1516.0

  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 \[ \frac{\mathrm{d}y}{ \mathrm{d}x} = \frac{10x - 4y}{ 3x - y}\;. \]

Show 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
2000 Paper 2 Q8
D: 1600.0 B: 1500.1

  1. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
  2. Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.
\noindent [The approximations, valid for small \(\theta\), \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-{\textstyle\frac12}\,\theta^2\) may be assumed.]

1995 Paper 2 Q8
D: 1600.0 B: 1500.8

If there are \(x\) micrograms of bacteria in a nutrient medium, the population of bacteria will grow at the rate \((2K-x)x\) micrograms per hour. Show that, if \(x=K\) when \(t=0\), the population at time \(t\) is given by \[ x(t)=K+K\frac{1-\mathrm{e}^{-2Kt}}{1+\mathrm{e}^{-2Kt}}. \] Sketch, for \(t\geqslant0\), the graph of \(x\) against \(t\). What happens to \(x(t)\) as \(t\rightarrow\infty\)? Now suppose that the situation is as described in the first paragraph, except that we remove the bacteria from the nutrient medium at a rate \(L\) micrograms per hour where \(K^{2}>L\). We set \(\alpha=\sqrt{K^{2}-L}.\) Write down the new differential equation for \(x\). By considering a new variable \(y=x-K+\alpha,\) or otherwise, show that, if \(x(0)=K\) then \(x(t)\rightarrow K+\alpha\) as \(t\rightarrow\infty\).

Show Solution
\begin{align*} && \dot{x} &= (2K-x)x \\ \Rightarrow && \int \d t &= \int \frac{1}{(2K-x)x} \d x \\ &&&= \int \frac1{2K}\left ( \frac{1}{2K-x} + \frac{1}{x} \right) \d x \\ &&&= \frac{1}{2K} \left (\ln x - \ln (2K-x) \right) \\ \Rightarrow && 2Kt+C &= \ln \frac{x}{2K-x} \\ t = 0, x = K: && C &= \ln \frac{K}{2K-K} = 0 \\ \Rightarrow && e^{2Kt} &= \frac{x}{2K-x} \\ \Rightarrow && e^{-2Kt} &= \frac{2K}{x} -1 \\ \Rightarrow && x &= \frac{2K}{1+e^{-2Kt}} \\ &&&= K + K \frac{1-e^{-2Kt}}{1+e^{-2Kt}} \end{align*}
TikZ diagram
As \(t \to \infty\) \(x(t) \to 2K\) We now have \[ \dot{x} = (2K-x)x - L\] Suppose \(y = x - K + \alpha\), where \(\alpha = \sqrt{K^2-L}\) then, \begin{align*} && \dot{y} &= \dot{x} \\ &&&= (2K-x)x - L \\ &&&= (2K - (y+K-\alpha))(y+K-\alpha) - L \\ &&&= (K+\alpha - y)y + (K+\alpha-y)(K-\alpha) - L \\ &&&= (K+\alpha-y)y + K^2-\alpha^2 -y(K-\alpha) - L \\ &&&= (K+\alpha-y)y -y(K-\alpha) \\ &&&= (2\alpha-y)y \end{align*} Note that when \(t = 0, x = K, y = \alpha\) so we have the original equation but with \(x \to y\) and \(\alpha \to K\), in particular as \(t \to \infty\) \(y \to 2\alpha\) and \(x \to 2\alpha - \alpha + K = K +\alpha\)
1990 Paper 1 Q7
D: 1500.0 B: 1500.0

Let \(y,u,v,P\) and \(Q\) all be functions of \(x\). Show that the substitution \(y=uv\) in the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q \] leads to an equation for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) in terms of \(x,Q\) and \(u\), provided that \(u\) satisfies a suitable first order differential equation. Hence or otherwise solve \[ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}}, \] given that \(y(1)=0\). For what set of values of \(x\) is the solution valid?

Show Solution
Suppose \(y = uv\) then and suppose \(\frac{\d u}{\d x} + P u = 0\) then \begin{align*} && \frac{\d y}{\d x} + Py &= Q \\ && uv' + u'v + Puv &= Q \\ && uv' &= Q \\ && \frac{\d v}{\d x} &= \frac{Q}{u} \end{align*} Consider \begin{align*} && 0 &= \frac{\d u}{\d x} - \frac{2u}{x+1} \\ \Rightarrow && \ln u &= 2\ln (1 + x) + C \\ \Rightarrow && u &= A(1+x)^2 \end{align*} and \begin{align*} && \frac{\d v}{\d x} &= \frac1{A}(x+1)^{\frac12} \\ \Rightarrow && v &= \frac2{3A}(x+1)^{\frac32} + k \\ \Rightarrow && y &= \frac23(x+1)^\frac72 + k(x+1)^2 \\ && 0 &= y(1) \\ &&&= \frac23 2^{7/2}+k2^2 \\ \Rightarrow && k &= -\frac{2^{5/2}}{3} \\ \Rightarrow && y &= \frac23 (x+1)^{7/2} - \frac{2^{5/2}}{3}(x+1)^2 \end{align*}
1990 Paper 3 Q8
D: 1700.0 B: 1484.7

Let \(P,Q\) and \(R\) be functions of \(x\). Prove that, for any function \(y\) of \(x\), the function \[ Py''+Qy'+Ry \] can be written in the form \(\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),\) where \(p\) and \(q\) are functions of \(x\), if and only if \(P''-Q'+R=0.\) Solve the differential equation \[ (x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x}, \] given that when \(x=2,y=2\mathrm{e}^{2}\) and \(y'=0.\)

Show Solution
Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}
1989 Paper 2 Q9
D: 1600.0 B: 1515.3

The matrix \(\mathbf{F}\) is defined by \[ \mathbf{F}=\mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}, \] where $\mathbf{A}=\begin{pmatrix}-3 & -1\\ 8 & 3 \end{pmatrix} \( and \) t \( is a variable scalar. Evaluate \)\mathbf{A}^{2}$, and show that \[ \mathbf{F}=\mathbf{I}\cosh t+\mathbf{A}\sinh t. \] Show also that \(\mathbf{F}^{-1}=\mathbf{I}\cosh t-\mathbf{A}\sinh t\), and that \(\dfrac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}=\mathbf{FA}\). The vector $\mathbf{r}=\begin{pmatrix}x(t)\\ y(t) \end{pmatrix}$ satisfies the differential equation \[ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}=\mathbf{0}, \] with \(x=\alpha\) and \(y=\beta\) at \(t=0.\) Solve this equation by means of a suitable matrix integrating factor, and hence show that \begin{alignat*}{1} x(t) & =\alpha\cosh t+(3\alpha+\beta)\sinh t\\ y(t) & =\beta\cosh t-(8\alpha+3\beta)\sinh t. \end{alignat*}

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\begin{align*} \begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}^2 &= \begin{pmatrix} 9-8 & 3-3 \\ -24+24 & -8+9 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \mathbf{I} \end{align*} Therefore: \begin{align*} \mathbf{F} &= \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n} \\ &= \mathbf{I} + \sum_{n=1}^{\infty} \frac{1}{(2n)!}t^{2n} \mathbf{I} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}t^{2n+1} \mathbf{A} \\ &= \cosh t \mathbf{I} + \sinh t \mathbf{A} \end{align*} Notice that \begin{align*} \mathbf{F} (\mathbf{I}\cosh t-\mathbf{A}\sinh t) &= (\mathbf{I}\cosh t+\mathbf{A}\sinh t)(\mathbf{I}\cosh t-\mathbf{A}\sinh t) \\ &= \mathbf{I}^2 \cosh^2 t+\mathbf{A}(\sinh t \cosh t - \cosh t \sinh t) - \mathbf{A}^2\sinh^2 t \\ &= \mathbf{I} \cosh^2 t - \mathbf{I} \sinh^2 t \\ &= \mathbf{I} \end{align*} Therefore \(\mathbf{F}^{-1} = \mathbf{I}\cosh t-\mathbf{A}\sinh t\) \begin{align*} \frac{\d \mathbf{F}}{\d t} &= \frac{\d }{\d t} \left [ \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}\right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^n \\ &= \left ( \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^{n-1} \right) \mathbf{A} \\ &= \mathbf{F}\mathbf{A} \end{align*} \begin{align*} && \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}&=\mathbf{0} \\ \Rightarrow && \mathbf{F} \frac{\d \mathbf{r}}{\d t} + \mathbf{FAr} &= \mathbf{0} \\ && \frac{\d }{\d t} \left ( \mathbf{F} \mathbf{r}\right) &= 0 \\ \Rightarrow && \mathbf{Fr} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= \mathbf{F}^{-1}\mathbf{c} \\ &&&= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\mathbf{c} \\ t = 0: && \binom{\alpha}{\beta} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\binom{\alpha}{\beta} \\ &&&= \binom{\alpha \cosh t}{\beta \cosh t} - \binom{-3\alpha-\beta}{8\alpha + 3\beta}\sinh t \\ &&&= \binom{\alpha \cosh t + (3\alpha + \beta) \sinh t}{\beta \cosh t -(8\alpha + 3\beta)\sinh t} \end{align*} as required
1987 Paper 1 Q3
D: 1500.0 B: 1500.0

By substituting \(y(x)=xv(x)\) in the differential equation \[ x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v}, \] or otherwise, find the solution \(v(x)\) that satisfies \(v=1\) when \(x=1\). What value does this solution approach when \(x\) becomes large?

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Let \(y = xv\) then \(y' = v + xv'\) and so \(x^2y' = x^2v + x^3v'\) Our differential equation is now: \begin{align*} && x^2 y' &= \frac{1+y^2}{(1+x^2)\frac{y}{x}} \\ \Rightarrow && xy' &= \frac{(1+y^2)}{(1+x^2)y} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x(1+x^2)} \\ \Rightarrow && \frac{y}{1+y^2} \frac{\d y}{\d x} &= \frac{1}{x} - \frac{x}{1+x^2} \\ \Rightarrow && \frac12 \ln(1+y^2) &= \ln x - \frac12 \ln(1+x^2) + C \\ \Rightarrow && \frac12 \ln (1 + y^2) &= \frac12 \ln \l \frac{x^2}{1+x^2}\r + C \\ \Rightarrow && 1+y^2 &= \frac{Dx^2}{1+x^2} \\ \Rightarrow && D &= 4 \quad \quad: (x = 1, v = 1, y = 1) \\ \Rightarrow && 1 + x^2v^2&= \frac{4x^2}{1+x^2}\\ \Rightarrow && v^2 &= \frac{3x^2-1}{x^2(1+x^2)} \\ \Rightarrow && v &= \sqrt{\frac{3x^2-1}{x^2(1+x^2)}} \\ \end{align*} As \(x \to \infty\), \(v \to 0\)