Separable variables - partial fractions

7 Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x \left( 3 y ^ { 2 } + 10 y + 3 \right)$$ obtaining an expression for \(y\) in terms of \(x\).

13. (a) Express \(\frac { 1 } { ( 4 - x ) ( 2 - x ) }\) in partial fractions. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation
where \(k\) is a constant.
(b) solve the differential equation and show that the solution can be written as $$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$ Given that \(k = 0.1\) (c) find the value of \(t\) when \(x = 1\), giving your answer, in seconds, to 3 significant figures. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$ where \(k\) is a constant. $$\text { Given that when } t = 0 , x = 0$$ (b) solve the differential equation and show that the solution can be written as

10. (a) Write \(\frac { 1 } { ( H - 5 ) ( H + 3 ) }\) in partial fraction form. The depth of water in a storage tank is being monitored.
The depth of water in the tank, \(H\) metres, is modelled by the differential equation $$\frac { \mathrm { d } H } { \mathrm {~d} t } = - \frac { ( H - 5 ) ( H + 3 ) } { 40 }$$ where \(t\) is the time, in days, from when monitoring began.
Given that the initial depth of water in the tank was 13 m ,
(b) solve the differential equation to show that $$H = \frac { 10 + 3 \mathrm { e } ^ { - 0.2 t } } { 2 - \mathrm { e } ^ { - 0.2 t } }$$ (c) Hence find the time taken for the depth of water in the tank to fall to 8 m .
(Solutions relying entirely on calculator technology are not acceptable.) According to the model, the depth of water in the tank will eventually fall to \(k\) metres.
(d) State the value of the constant \(k\).

1. During a chemical reaction, a compound is being made from two other substances.

At time \(t\) hours after the start of the reaction, \(x \mathrm {~g}\) of the compound has been produced. Assuming that \(x = 0\) initially, and that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$

1. show that it takes approximately 7 minutes to produce 2 g of the compound.
2. Explain why it is not possible to produce 3 g of the compound.

8. The diagram shows a hemispherical bowl of radius 5 cm . The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time \(t\) minutes, the depth of water is \(h \mathrm {~cm}\) and the volume of water in the bowl is \(V \mathrm {~cm} ^ { 3 }\), where $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 15 - h ) .$$ In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to \(V\).

1. Show that $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { k h ( 15 - h ) } { 3 ( 10 - h ) } ,$$ where \(k\) is a positive constant.
2. Express \(\frac { 3 ( 10 - h ) } { h ( 15 - h ) }\) in partial fractions. Given that when \(t = 0 , h = 5\),
3. show that $$h ^ { 2 } ( 15 - h ) = 250 \mathrm { e } ^ { - k t } .$$ Given also that when \(t = 2 , h = 4\),
4. find the value of \(k\) to 3 significant figures.

7

1. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
   [0pt] [3]
2. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.

1 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( x + 1 ) }\), given that when \(x = 1 , y = 1\). Your answer should express \(y\) explicitly in terms of \(x\).

4 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?

8

1. Find the quotient and the remainder when \(x ^ { 2 } - 5 x + 6\) is divided by \(x - 1\).
2. (a) Find the general solution of the differential equation $$\left( \frac { x - 1 } { x ^ { 2 } - 5 x + 6 } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y - 5 .$$ (b) Given that \(y = 7\) when \(x = 8\), find \(y\) when \(x = 6\).

8 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
4. What does this solution indicate about the long-term height of the tree?
5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?

12 Find the general solution of the differential equation \(\left( 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 20 x - 35 )\).
Give your answer in the form \(y = \mathrm { f } ( x )\). \section*{END OF QUESTION PAPER} \section*{OCR
Oxford Cambridge and RSA}

16 In this question you must show detailed reasoning. Find the particular solution of the differential equation $$\frac { d y } { d x } = \frac { 9 y } { ( x - 1 ) ( x + 2 ) }$$ given that \(x = 2\) when \(y = 16\). \section*{END OF QUESTION PAPER}

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. \(\_\_\_\_\)

10

1. Using partial fractions, find the general solution of the differential equation $$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = y - y ^ { 3 } \text { for } 0 < y < 1$$ giving your solution in the form \(y = \mathrm { f } ( x )\).
2. Determine \(\lim _ { x \rightarrow - \infty } \mathrm { f } ( x )\) and \(\lim _ { x \rightarrow + \infty } \mathrm { f } ( x )\).

Solve the differential equation \(\frac{dy}{dx} = \frac{y}{x(x+1)}\), given that when \(x = 1\), \(y = 1\). Your answer should express \(y\) explicitly in terms of \(x\). [8]

\includegraphics{figure_2} Figure 2 shows a hemispherical bowl of radius 5 cm. The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time \(t\) minutes, the depth of water is \(h\) cm and the volume of water in the bowl is \(V\) cm³, where $$V = \frac{1}{3}\pi h^2(15 - h).$$ In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to \(V\).

1. Show that $$\frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)},$$ where \(k\) is a positive constant. [5]
2. Express \(\frac{3(10-h)}{h(15-h)}\) in partial fractions. [3]

Given that when \(t = 0\), \(h = 5\),

1. show that $$h^2(15-h) = 250e^{-kt}.$$ [6]

Given also that when \(t = 2\), \(h = 4\),

1. find the value of \(k\) to 3 significant figures. [3]

The points \((x, y)\) on the curve \(C\) satisfy \((x + 1)(x + 2) \frac{dy}{dx} = xy\). The line with equation \(y = 2x + 5\) is the tangent to \(C\) at a point \(P\).

1. Find the coordinates of \(P\). [4]
2. Find the equation of \(C\), giving your answer in the form \(y = f(x)\). [8]

In a chemical reaction, the mass \(m\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dm}{dt} = \frac{m}{t(1 + 2t)}.$$ At time 1 minute, the mass of the chemical is 1 gram.

1. Solve the differential equation to show that \(m = \frac{3t}{(1 + 2t)}\). [8]
2. Hence
   1. find the time when the mass is 1.25 grams, [2]
   2. show what happens to the mass of the chemical as \(t\) becomes large. [2]

Find the general solution of the differential equation $$(2x^3 - 3x^2 - 11x + 6)\frac{dy}{dx} = y(20x - 35).$$ Give your answer in the form \(y = f(x)\). [9]