4.10a General/particular solutions: of differential equations

3 This question concerns the family of differential equations \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - x ^ { a } y \left( { } ^ { * } \right)\) where \(a\) is \(- 1,0\) or 1 .

1. Determine and describe geometrically the isoclines of (\textit{) when
   1. \(a = - 1\),
   2. \(a = 0\),
   3. \(a = 1\).
2. In this part of the question \(a = 0\).
   1. Write down the solution to \(( * )\) which passes through the point \(( 0 , b )\) where \(b \neq 1\).
   2. Write down the equation of the asymptote to this solution.
3. In this part of the question \(a = - 1\).
   1. Write down the solution to \(( * )\) which passes through the point \(( c , d )\) where \(c \neq 0\).
   2. Describe the relationship between \(c\) and \(d\) when the solution in part (i) has a stationary point.
4. In this part of the question \(a = 1\).
   1. The standard Runge-Kutta method of order 4 for the solution of the differential equation \(\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }\) is as follows. \(k _ { 1 } = h \mathrm { f } \left( x _ { n } , y _ { n } \right)\) \(k _ { 2 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 1 } } { 2 } \right)\) \(k _ { 3 } = h \mathrm { f } \left( x _ { n } + \frac { h } { 2 } , y _ { n } + \frac { k _ { 2 } } { 2 } \right)\) \(k _ { 4 } = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 3 } \right)\) \(y _ { n + 1 } = y _ { n } + \frac { 1 } { 6 } \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right)\).
      Construct a spreadsheet to solve (}) in the case \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\). State the formulae you have used in your spreadsheet.
   2. Use your spreadsheet with \(h = 0.05\) to find an approximation to the value of \(y\) when \(x = 1\).
   3. The solution to \(( * )\) in which \(x _ { 0 } = 0\) and \(y _ { 0 } = 1.5\) has a maximum point ( \(r , s\) ) with \(0 < r < 1\). Use your spreadsheet with suitable values of \(h\) to estimate \(r\) to two decimal places. Justify your answer.

3 In this question you are required to consider the family of differential equations \(\frac { d y } { d x } = \frac { y ^ { a } } { x + 1 } - \frac { 1 } { y } ( * )\) and its solutions. The parameter \(a\) is a real number. You should assume that \(x \geqslant 0\) and \(y > 0\) throughout this question.

1. In this part of the question \(a = 1\).
   1. On the axes in the Printed Answer Booklet
      • Sketch the isocline defined by \(\frac { d y } { d x } = 0\).
      • Shade and label the region in which \(\frac { \mathrm { dy } } { \mathrm { dx } } > 0\).
      • Shade and label the region in which \(\frac { \mathrm { dy } } { \mathrm { dx } } < 0\).
      • For \(b > 0\), find, in terms of \(b\), the solution to \(( * )\) which passes through the point \(( 0 , b )\).
      • Determine
      • The values of \(b > 0\) for which the solution in (ii) has a turning point.
      • The corresponding maximum value of \(y\).
      • Fig. 3.1 and Fig. 3.2 show tangent fields for two distinct but unspecified values of \(a\). In each case a sketch of the solution curve \(y = \mathrm { g } ( x )\) which passes through \(( 0,2 )\) is shown for \(0 \leqslant x \leqslant 0.5\).
      \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{43fdb360-0f80-4794-917c-f28b04181fa4-4_656_648_1777_301} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
      \end{figure} \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{43fdb360-0f80-4794-917c-f28b04181fa4-4_656_652_1777_1117} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
      \end{figure}
      1. For the case in Fig. 3.1 suggest a possible value of \(a\).
      2. For the case in Fig. 3.2 suggest a possible value of \(a\).
      3. In each case, continue the sketch of the solution curves for \(0.5 \leqslant x \leqslant 5\) in the Printed Answer Booklet.
      4. State a feature which is present in one of the curves in part (iii) for \(0.5 \leqslant x \leqslant 5\) but not in the other.
2. 1. The Euler method for the solution of the differential equation \(\frac { \mathrm { dy } } { \mathrm { dx } } = \mathrm { f } ( x , y )\) is as follows $$y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } \right)$$ It is given that \(x _ { 0 } = 0\) and \(y _ { 0 } = 2\).
      Use your spreadsheet with \(h = 0.1\) to approximate the value of \(y\) when \(x = 3\) for the solution to (*) in which \(y = 2\) when \(x = 0\).
   2. In this part of the question \(a = - 0.2\). Use your spreadsheet to approximate, to \(\mathbf { 1 }\) decimal place, the \(x\)-coordinate of the local maximum for the solution to (*) in which \(y = 2\) when \(x = 0\).

4 In this question you are required to consider the family of differential equations $$\frac { d P } { d t } = r P \left( 1 - \frac { P } { K } \right) , \quad t \geqslant 0 , \quad P ( t ) \geqslant 0 \left( ^ { * } \right)$$ where \(r\) and \(K\) are positive constants. This differential equation can be used as a model for the size of a population \(P\) as a function of time \(t\).

1. 1. Determine the values of \(P\) for which
      $$\frac { d P } { d t } = 2 P ^ { 1.25 } \left( 1 - \frac { P } { 1000 } \right) ^ { 1.5 } , t \geqslant 0 , P ( t ) \geqslant 0 ( * * )$$ The diagram shows the tangent field to (**), and a solution in which \(P = 1\) when \(t = 0\), produced using a much more accurate numerical method. \includegraphics[max width=\textwidth, alt={}, center]{4715d0f0-a860-4189-802f-1d2d019e1115-4_899_1552_1763_319}
      1. The Euler method for the solution of the differential equation \(f ( t , P ) = \frac { d P } { d t }\) is as follows $$P _ { n + 1 } = P _ { n } + h f \left( t _ { n } , P _ { n } \right)$$ It is given that \(t _ { 0 } = 0\) and \(P _ { 0 } = 1\).

3 This question concerns the family of differential equations $$\frac { d y } { d x } = x ^ { 2 } - y + \operatorname { acos } ( x ) \cos ( y ) \quad ( * * )$$ where \(a\) is a constant, \(x \geqslant 0\) and \(y > 0\).

1. In this part of the question \(a = 0\).
   1. Find the solution to (\textbf{) in which \(y = 1\) when \(x = 0\).
   2. In this part of the question \(m\) is a real number. Show that the equation of the isocline \(\frac { \mathrm { dy } } { \mathrm { dx } } = \mathrm { m }\) is a parabola.
   3. Using the result given in part (a)(ii), or otherwise, sketch the tangent field for (}) on the axes in the Printed Answer Booklet.
2. Fig. 3.1 and Fig. 3.2 show the tangent fields for two distinct and unspecified values of \(a\). In each case, a sketch of the solution curve \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) which passes through the point \(( 0,2 )\) is shown for \(0 \leqslant x \leqslant \frac { 1 } { 2 }\). \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 3.1} \includegraphics[alt={},max width=\textwidth]{6d485052-b0db-4c33-b374-4fd7b6f0759c-4_399_666_1324_317}
   \end{figure} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 3.2} \includegraphics[alt={},max width=\textwidth]{6d485052-b0db-4c33-b374-4fd7b6f0759c-4_397_661_1324_1192}
   \end{figure}
   1. In each case, continue the sketch of the solution curve for \(\frac { 1 } { 2 } \leqslant x \leqslant 3\) on the axes in the Printed Answer Booklet.
   2. State one feature which is present in the continued solution curve for Fig. 3.1 that is not a feature of the continued solution curve for Fig. 3.2.
   3. Using a slider for \(a\), or otherwise, estimate the value of \(a\) for the solution curve shown in Fig. 3.2.
3. The Euler method for the solution of the differential equation \(\frac { d y } { d x } = f ( x , y )\) is as follows. $$\begin{aligned} & y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } \right) \\ & x _ { n + 1 } = x _ { n } + h \end{aligned}$$
   1. Construct a spreadsheet to solve (), so that the value of \(a\) and the value of \(h\) can be varied, in the case \(x _ { 0 = 0\) and \(y _ { 0 } = 1\). State the formulae you have used in your spreadsheet.
   2. In this part of the question \(a = 0\). Use your spreadsheet with \(h = 0.1\) to approximate the value of \(y\) when \(x = 0.5\) for the solution to (}) in which \(y = 1\) when \(x = 0\).
   3. Using part (a)(i), state the accuracy of the approximate value of \(y\) given in part (c)(ii).
   4. State one change to your spreadsheet that could improve the accuracy of the approximate value of \(y\) found in part (c)(ii).
4. The modified Euler method for the solution of the differential equation \(\frac { d y } { d x } = f ( x , y )\) is as follows. \(k _ { 1 } = h f \left( x _ { n } , y _ { n } \right)\) \(k _ { 2 } = h f \left( x _ { n } + h , y _ { n } + k _ { 1 } \right)\) \(y _ { n + 1 } = y _ { n } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)\) \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { x } _ { \mathrm { n } } + \mathrm { h }\)
   1. Adapt your spreadsheet from part (c)(i) to a spreadsheet to solve (**), so that the value of \(a\) and the value of \(h\) can be varied, in the case \(x _ { 0 } = 0\) and \(y _ { 0 } = 1\). State the formulae you have used in your spreadsheet.
   2. In this part of the question \(a = - 0.5\). Use the spreadsheet from part (d)(i) with \(h = 0.1\) to approximate the value of \(y\) when \(x = 0.5\) for the solution to \(( * * )\) in which \(y = 1\) when \(x = 0\). In this part of the question \(a = - 0.5\). The solution to (**) in which \(y = 1\) when \(x = 0\) has a turning point with coordinates \(( c , d )\) where \(0 < c < 1\).
   3. Use the spreadsheet in part (d)(i) to determine the value of \(c\) correct to \(\mathbf { 1 }\) decimal place.
   4. Use the spreadsheet in part (d)(i) to determine the value of \(d\) correct to \(\mathbf { 3 }\) decimal places.

3 This question explores the family of differential equations \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { 1 + a x + 2 y }\) for various values of the parameter \(a\). Fig. 3 shows the tangent field in the case \(a = 1\). \begin{figure}[h] \end{figure}

1. (A) Sketch the tangent field in the case \(a = - 2\).
   (B) Explain why the tangent field is not defined for the whole coordinate plane.
   (C) Give an inequality which describes the region in which the tangent field is defined.
   (D) Find a value of \(a\) such that the region for which the tangent field is defined includes the entire \(x\)-axis.
2. (A) For the case \(a = 1\), with \(y = 1\) when \(x = 0\), construct a spreadsheet for the Runge-Kutta method of order 2 with formulae as follows, where \(\mathrm { f } ( x , y ) = \frac { \mathrm { d } y } { \mathrm {~d} x }\). $$\begin{aligned} k _ { 1 } & = h \mathrm { f } \left( x _ { n } , y _ { n } \right) \\ k _ { 2 } & = h \mathrm { f } \left( x _ { n } + h , y _ { n } + k _ { 1 } \right) \\ y _ { n + 1 } & = y _ { n } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right) \end{aligned}$$ State the formulae you have used in your spreadsheet.
   (B) Use your spreadsheet to obtain the value of \(y\) correct to 4 decimal places when \(x = 1\) for
3. (A) For the case \(a = 0\) find the analytical solution that passes through the point ( 0,1 ).
   (B) Verify that the solution in part (iii) (A) is a solution to the differential equation.
   (C) Use the solution in part (iii) (A) to find the value of \(y\) correct to 4 decimal places when \(x = 1\).
4. (A) Verify that \(y = - \frac { a } { 2 } x + \frac { a ^ { 2 } } { 8 } - \frac { 1 } { 2 }\) is a solution for all cases when \(a \leq 0\).
   (B) Show that this is the only straight line solution in these cases. \section*{Copyright Information:} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
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1. The temperature, \(\theta ^ { \circ } \mathrm { C }\), of coffee in a cup, \(t\) minutes after the cup of coffee is put in a room, is modelled by the differential equation

$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 )$$ where \(k\) is a constant.
The coffee has an initial temperature of \(80 ^ { \circ } \mathrm { C }\) Using \(k = 0.1\)

1. use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { 0 } = \frac { y _ { 1 } - y _ { 0 } } { h }\) to estimate the temperature of the coffee 3 minutes after it was put in the room. The coffee in a different cup, which also had an initial temperature of \(80 ^ { \circ } \mathrm { C }\) when it was put in the room, cools more slowly.
2. Use this information to suggest how the value of \(k\) would need to be changed in the model.

   V349 SIHI NI IMIMM ION OC

   VJYV SIHIL NI LIIIM ION OO

   VJYV SIHIL NI JIIYM ION OC

1. Julie decides to start a business breeding rabbits to sell as pets.

Initially she buys 20 rabbits. After \(t\) years the number of rabbits, \(R\), is modelled by the differential equation $$\frac { \mathrm { d } R } { \mathrm {~d} t } = 2 R + 4 \sin t \quad t > 0$$ Julie needs to have at least 40 rabbits before she can start to sell them.
Use two iterations of the approximation formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$ to find out if, according to the model, Julie will be able to start selling rabbits after 4 months.

1. The velocity \(v \mathrm {~ms} ^ { - 1 }\), of a raindrop, \(t\) seconds after it falls from a cloud, is modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.1 v ^ { 2 } + 10 \quad t \geqslant 0$$ Initially the raindrop is at rest.

1. Use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }\) to estimate the velocity of the raindrop 1 second after it falls from the cloud. Given that the initial acceleration of the raindrop is found to be smaller than is suggested by the current model,
2. refine the model by changing the value of one constant.

4. $$\left[ \begin{array} { l } \text { The Taylor series expansion of } \mathrm { f } ( x ) \text { about } x = a \text { is given by } \\ \mathrm { f } ( x ) = \mathrm { f } ( a ) + ( x - a ) \mathrm { f } ^ { \prime } ( a ) + \frac { ( x - a ) ^ { 2 } } { 2 ! } \mathrm { f } ^ { \prime \prime } ( a ) + \ldots + \frac { ( x - a ) ^ { r } } { r ! } \mathrm { f } ^ { ( r ) } ( a ) + \ldots \end{array} \right]$$ The curve with equation \(y = \mathrm { f } ( x )\) satisfies the differential equation $$\cos x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \sin x = 0$$ Given that \(\left( \frac { \pi } { 4 } , 1 \right)\) is a stationary point of the curve,

1. determine the nature of this stationary point, giving a reason for your answer.
2. Show that \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \sqrt { 2 } - 2\) at this stationary point.
3. Hence determine a series solution for \(y\), in ascending powers of \(\left( x - \frac { \pi } { 4 } \right)\) up to and including the term in \(\left( x - \frac { \pi } { 4 } \right) ^ { 3 }\), giving each coefficient in simplest form.

1. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation

$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t ( t + 1 ) \frac { \mathrm { d } x } { \mathrm {~d} t } + 2 ( t + 1 ) x = 8 t ^ { 3 } \mathrm { e } ^ { t }$$ where \(P\) has displacement \(x\) metres from the origin \(O\) at time \(t\) minutes, \(t > 0\)

1. Show that the transformation \(x = t u\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} u } { \mathrm {~d} t } = 8 \mathrm { e } ^ { t }$$ Given that \(P\) is at \(O\) when \(t = \ln 3\) and when \(t = \ln 5\)
2. determine the particular solution of the differential equation (I)

1. A vibrating spring, fixed at one end, has an external force acting on it such that the centre of the spring moves in a straight line. At time \(t\) seconds, \(t \geqslant 0\), the displacement of the centre \(C\) of the spring from a fixed point \(O\) is \(x\) micrometres.

The displacement of \(C\) from \(O\) is modelled by the differential equation $$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 2 + t ^ { 2 } \right) x = t ^ { 4 }$$

1. Show that the transformation \(x = t v\) transforms equation (I) into the equation $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} t ^ { 2 } } + v = t$$
2. Hence find the general equation for the displacement of \(C\) from \(O\) at time \(t\) seconds.
3. 1. State what happens to the displacement of \(C\) from \(O\) as \(t\) becomes large.
   2. Comment on the model with reference to this long term behaviour.

4. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 0$$

1. Show that $$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = a x \frac { \mathrm {~d} ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } + b \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$$ where \(a\) and \(b\) are integers to be found.
2. Hence find a series solution, in ascending powers of \(x\), as far as the term in \(x ^ { 5 }\), of the differential equation (I) where \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)

9 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term? RECOGNISING ACHIEVEMENT \section*{ADVANCED GCE} \section*{4754/01B} \section*{MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) Paper B: Comprehension
   WEDNESDAY 21 MAY 2008
   Afternoon
   Time: Up to 1 hour
   Additional materials: Rough paper
   MEI Examination Formulae and Tables (MF 2) \section*{Candidate Forename}
   \includegraphics[max width=\textwidth, alt={}]{8ad99e2a-4cef-40b3-af8d-673b97536227-05_125_547_986_516}
   This document consists of \(\mathbf { 6 }\) printed pages, \(\mathbf { 2 }\) blank pages and an insert. 1 Complete these Latin square puzzles.

6. 2	1	3
   3

7. \includegraphics[max width=\textwidth, alt={}, center]{8ad99e2a-4cef-40b3-af8d-673b97536227-06_391_419_836_854} 2 In line 51, the text says that the Latin square

   1	2	3	4
   3	1	4	2
   2	4	1	3
   4	3	2	1

   could not be the solution to a Sudoku puzzle.
   Explain this briefly.
   3 On lines 114 and 115 the text says "It turns out that there are 16 different ways of filling in the remaining cells while keeping to the Sudoku rules. One of these ways is shown in Fig.10." Complete the grid below with a solution different from that given in Fig. 10.

   1	2	3	4

   4 Lines 154 and 155 of the article read "There are three other embedded Latin squares in Fig. 14; one of them is illustrated in Fig. 16." Indicate one of the other two embedded Latin squares on this copy of Fig. 14.

   4	2	3	1
   		2	4
   		4	2
   2	4	1	3

   5 The number of \(9 \times 9\) Sudokus is given in line 121 .
   Without doing any calculations, explain why you would expect 9! to be a factor of this number.
   6 In the table below, \(M\) represents the maximum number of givens for which a Sudoku puzzle may have no unique solution (Investigation 3 in the article). \(s\) is the side length of the Sudoku grid and \(b\) is the side length of its blocks.

   Block side length, \(b\)	Sudoku, \(s \times s\)	\(M\)
   1	\(1 \times 1\)	-
   2	\(4 \times 4\)	12
   3	\(9 \times 9\)
   4	\(16 \times 16\)
   5

8. Complete the table.
9. Give a formula for \(M\) in terms of \(b\).
   7 A man is setting a Sudoku puzzle and starts with this solution.

   1	2	3	4	5	6	7	8	9
   4	5	6	8	9	7	3	1	2
   7	8	9	3	1	2	5	6	4
   2	3	1	5	6	4	8	9	7
   5	6	4	9	7	8	1	2	3
   8	9	7	1	2	3	6	4	5
   3	1	2	6	4	5	9	7	8
   6	4	5	7	8	9	2	3	1
   9	7	8	2	3	1	4	5	6

   He then removes some of the numbers to give the puzzles in parts (i) and (ii). In each case explain briefly, and without trying to solve the puzzle, why it does not have a unique solution.
   [0pt] [2,2]

10. 1	2		4		6			9
    4			8	9			1
    	8						6
    2		1			4			7
    	6	4		7	8	1	2
    8	9			2			4
    	1		6	4		9	7
    6	4		7		9			1
    9		8	2		1	4		6

11. 1	2	3	4	5	6	7	8	9
    4	5	6	8	9	7	3	1	2
    7	8	9				5	6	4
    2	3	1	5	6	4	8	9	7
    5	6	4	9	7	8	1	2	3
    8	9	7				6	4	5
    3	1	2	6	4	5	9	7	8
    6	4	5	7	8	9	2	3	1
    9	7	8				4	5	6

12. \(\_\_\_\_\)
13. \(\_\_\_\_\)

7 The diagram shows a curve which starts from the point \(A\) with coordinates ( 0,2 ). The curve is such that, at every point \(P\) on the curve, $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } s$$ where \(s\) is the length of the \(\operatorname { arc } A P\). \includegraphics[max width=\textwidth, alt={}, center]{587aac5c-fbc2-41d2-b1b3-16f3f7851d9d-4_399_764_1324_605}

1. 1. Show that $$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \sqrt { 4 + s ^ { 2 } }$$ (3 marks)
   2. Hence show that $$s = 2 \sinh \frac { x } { 2 }$$
   3. Hence find the cartesian equation of the curve.
2. Show that $$y ^ { 2 } = 4 + s ^ { 2 }$$

3

1. Show that \(y = x ^ { 3 } - x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$
2. By differentiating \(\left( x ^ { 2 } - 1 \right) y = c\) implicitly, where \(y\) is a function of \(x\) and \(c\) is a constant, show that \(y = \frac { c } { x ^ { 2 } - 1 }\) is a solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 0$$
3. Hence find the general solution of $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$

2 A particle \(P\) of mass 4.5 kg is free to move along the \(x\)-axis. In a model of the motion it is assumed that \(P\) is acted on by two forces:

• a constant force of magnitude \(f \mathrm {~N}\) in the positive \(x\) direction;
• a resistance to motion, \(R \mathrm {~N}\), whose magnitude is proportional to the speed of \(P\).

At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). When \(t = 0 , P\) is at the origin \(O\) and is moving in the positive direction with speed \(u \mathrm {~ms} ^ { - 1 }\), and when \(v = 5 , R = 2\). \begin{enumerate}[label=(\alph*)] \item Show that, according to the model, \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 10 f - 4 v } { 45 }\). \item

1. By solving the differential equation in part (a), show that \(v = \frac { 1 } { 2 } \left( 5 f - ( 5 f - 2 u ) \mathrm { e } ^ { - \frac { 4 } { 45 } t } \right)\).
2. Describe briefly how, according to the model, the speed of \(P\) varies over time in each of the following cases.
   The flat surface of a smooth solid hemisphere of radius \(r\) is fixed to a horizontal plane on a planet where the acceleration due to gravity is denoted by \(\gamma\). \(O\) is the centre of the flat surface of the hemisphere. A particle \(P\) is held at a point on the surface of the hemisphere such that the angle between \(O P\) and the upward vertical through \(O\) is \(\alpha\), where \(\cos \alpha = \frac { 3 } { 4 }\). \(P\) is then released from rest. \(F\) is the point on the plane where \(P\) first hits the plane (see diagram).
   1. Find an exact expression for the distance \(O F\). The acceleration due to gravity on and near the surface of the planet Earth is roughly \(6 \gamma\).
   2. Explain whether \(O F\) would increase, decrease or remain unchanged if the action were repeated on the planet Earth.

3 The resistive force, \(F\), on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, \(r\), the velocity of the sphere, \(v\), and the viscosity of the fluid, \(\eta\). You are given that \(\eta\) is measured in \(\mathrm { Nm } ^ { - 2 } \mathrm {~s}\).

1. By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: \(F = 6 \pi \eta ^ { \alpha } r ^ { \beta } v ^ { \gamma }\).
2. Explain whether or not it is possible to use dimensional analysis to verify that the constant \(6 \pi\) is correct.
3. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\). A sphere of radius \(r\) and mass \(m\) falls vertically from rest through the fluid. After a time \(t\) its velocity is \(v\).
4. By setting up and solving a differential equation, show that \(\mathrm { e } ^ { - k t } = \frac { g - k v } { g }\) where \(k = \frac { 6 \pi \eta r } { m }\). As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
5. Find, in terms of \(g\) and \(k\), the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
6. Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.

9

1. Show that the substitution \(u = \frac { 1 } { y ^ { 3 } }\) transforms the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\) into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x .$$
2. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\), given that \(y = \frac { 1 } { 2 }\) when \(x = 0\). Give your answer in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

7 The function \(f\) satisfies the differential equation $$x ^ { 2 } \mathrm { f } ^ { \prime \prime } ( x ) + ( 2 x - 1 ) \mathrm { f } ^ { \prime } ( x ) - 2 \mathrm { f } ( x ) = 3 \mathrm { e } ^ { x - 1 } + 1 ,$$ and the conditions \(f ( 1 ) = 2 , f ^ { \prime } ( 1 ) = 3\).

1. Determine \(f ^ { \prime \prime } ( 1 )\).
2. Differentiate (*) with respect to \(x\) and hence evaluate \(\mathrm { f } ^ { \prime \prime \prime } ( 1 )\).
3. Hence determine the Taylor series approximation for \(\mathrm { f } ( x )\) about \(x = 1\), up to and including the term in \(( x - 1 ) ^ { 3 }\).
4. Deduce, to 3 decimal places, an approximation for \(\mathrm { f } ( 1.1 )\).

5 The variables \(y\) and \(x\) are related by the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } , \quad - 2 < x < 2 .$$ By writing \(u = \frac { \mathrm { d } y } { \mathrm {~d} x }\), determine \(y\) explicitly in terms of \(x\), given that \(y = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }\) when \(x = 0\).

A train is moving along a straight horizontal track. At time t seconds, its velocity is \(v \mathrm {~ms} ^ { - 1 }\), its acceleration is \(a \mathrm {~ms} ^ { - 2 }\), and \(a\) is inversely proportional to V . At time \(\mathrm { t } = 1\), \(v = 5\) and \(a = 1 \cdot 8\). a) i) Write down a differential equation satisfied by V.
ii) Show that \(v ^ { 2 } = 18 t + 7\).
b) Find the time at which the magnitude of the velocity is equal to the magnitude of the acceleration. \section*{END OF PAPER}

A particle \(P\) of mass 0.25 kg moves in a straight line on a smooth horizontal surface. \(P\) starts at the point \(O\) with speed \(10 \text{ m s}^{-1}\) and moves towards a fixed point \(A\) on the line. At time \(t\) s the displacement of \(P\) from \(O\) is \(x\) m and the velocity of \(P\) is \(v \text{ m s}^{-1}\). A resistive force of magnitude \((5 - x)\) N acts on \(P\) in the direction towards \(O\).

1. Form a differential equation in \(v\) and \(x\). By solving this differential equation, show that \(v = 10 - 2x\). [6]
2. Find \(x\) in terms of \(t\), and hence show that the particle is always less than 5 m from \(O\). [5]

A particle \(P\) of mass \(0.1\) kg moves with decreasing speed in a straight line on a smooth horizontal surface. A horizontal resisting force of magnitude \(0.2e^{-x}\) N acts on \(P\), where \(x\) m is the displacement of \(P\) from a fixed point \(O\) on the line. The velocity of \(P\) is \(v\) m s\(^{-1}\) when its displacement from \(O\) is \(x\) m.

1. Show that $$v\frac{dv}{dx} = ke^{-x},$$ where \(k\) is a constant to be found. [2]

\(P\) passes through \(O\) with velocity \(2.2\) m s\(^{-1}\).

1. Calculate the value of \(x\) at the instant when the velocity of \(P\) is \(2\) m s\(^{-1}\). [4]
2. Show that the speed of \(P\) does not fall below \(0.917\) m s\(^{-1}\), correct to \(3\) significant figures. [2]

A particle \(P\) of mass \(0.4 \text{ kg}\) is released from rest at a point \(O\) on a smooth plane inclined at \(30°\) to the horizontal. When the displacement of \(P\) from \(O\) is \(x \text{ m}\) down the plane, the velocity of \(P\) is \(v \text{ ms}^{-1}\). A force of magnitude \(0.8e^{-x} \text{ N}\) acts on \(P\) up the plane along the line of greatest slope through \(O\).

1. Show that \(v \frac{dv}{dx} = 5 - 2e^{-x}\). [2]
2. Find \(v\) when \(x = 0.6\). [4]