First order differential equations (integrating factor)

11 Solve the differential equation \(\cosh x \frac { d y } { d x } - 2 y \sinh x = \cosh x\), given that \(y = 1\) when \(x = 0\).

1 Find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { y } { x } = x ,$$ giving \(y\) in terms of \(x\) in your answer.

Given the differential equation $$\sec x \frac{\mathrm{d}y}{\mathrm{d}x} + y\cos \sec x = 2$$ and \(x = \frac{\pi}{2}\) when \(y = 3\), find the value of \(y\) when \(x = \frac{\pi}{4}\). [8]

Find the general solution of the differential equation $$\frac{dy}{dx} = \frac{x}{x(1 + x^2)}$$ giving your answer in the form \(y = f(x)\). [10]

5 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x + y }$$ and \(y = 0\) when \(x = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

5.In this question u and v are functions of \(x\) .Given that \(\int \mathrm { u } \mathrm { d } x , \int \mathrm { v } \mathrm { d } x\) and \(\int \mathrm { uv } \mathrm { d } x\) satisfy $$\int \text { uv } \mathrm { d } x = \left( \int \mathrm { u } \mathrm {~d} x \right) \times \left( \int \mathrm { v } \mathrm {~d} x \right) \quad \text { uv } \neq 0$$

1. show that \(1 = \frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } + \frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) Given also that \(\frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } = \mathrm { sin } ^ { 2 } x\),
2. use part(a)to write down an expression,in terms of \(x\) ,for \(\frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) ,
3. show that $$\frac { 1 } { \mathrm { u } } \frac { \mathrm { du } } { \mathrm {~d} x } = \frac { 1 - 2 \sin x \cos x } { \sin ^ { 2 } x }$$
4. hence use integration to show that \(\mathrm { u } = A \mathrm { e } ^ { - \cot x } \operatorname { cosec } ^ { 2 } x\) ,where \(A\) is an arbitrary constant.
5. By differentiating \(\mathrm { e } ^ { \tan x }\) find a similar expression for v .

At time \(t = 0\), the position vector of a particle \(P\) is \(-3j\) m. At time \(t\) seconds, the position vector of \(P\) is \(\mathbf{r}\) metres and the velocity of \(P\) is \(\mathbf{v}\) m s\(^{-1}\). Given that $$\mathbf{v} - 2\mathbf{r} = 4e^t \mathbf{j},$$ find the time when \(P\) passes through the origin. [7]

7. A population \(P\) is growing at a rate which is modelled by the differential equation $$\frac { d P } { d t } - 0.1 P = 0.05 t$$ where \(t\) years is the time that has elapsed from the start of observations.
It is given that the population is 10000 at the start of the observations.

1. Solve the differential equation to obtain an expression for \(P\) in terms of \(t\).
2. Show that the population doubles between the sixth and seventh year after the observations began.
   (2)

8 A surge in the current, \(I\) units, through an electrical component at a time, \(t\) seconds, is to be modelled. The surge starts when \(t = 0\) and there is initially no current through the component. When the current has surged for 1 second it is measured as being 5 units. While the surge is occurring, \(I\) is modelled by the following differential equation. \(\left( 2 t - t ^ { 2 } \right) \frac { d l } { d t } = \left( 2 t - t ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - 2 ( t - 1 ) l\)

1. By using an integrating factor show that, according to the model, while the surge is occurring, \(I\) is given by \(\mathrm { I } = \left( 2 \mathrm { t } - \mathrm { t } ^ { 2 } \right) \left( \sin ^ { - 1 } ( \mathrm { t } - 1 ) + 5 \right)\). The surge lasts until there is again no current through the component.
2. Determine the length of time that the surge lasts according to the model.
3. Determine, according to the model, the rate of increase of the current at the start of the surge. Give your answer in an exact form.

1. (a) Show that the substitution \(y ^ { 2 } = \frac { 1 } { t }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y = x y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } t } { \mathrm {~d} x } - 2 t = - 2 x$$ (b) Solve differential equation (II) and determine \(y ^ { 2 }\) in terms of \(x\).

1. (a) Show that the transformation \(y = \frac { 1 } { z }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 2 y ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { z } { x } = - \frac { 2 } { x ^ { 2 } }$$ (b) Solve differential equation (II) to determine \(z\) in terms of \(x\).
(c) Hence determine the particular solution of differential equation (I) for which \(y = - \frac { 3 } { 8 }\) at \(x = 3\) Give your answer in the form \(y = \mathrm { f } ( x )\).

1. (a) For all the values of \(x\) where the identity is defined, prove that

$$\cot 2 x + \tan x \equiv \operatorname { cosec } 2 x$$ (b) Show that the substitution \(y ^ { 2 } = w \sin 2 x\), where \(w\) is a function of \(x\), transforms the differential equation $$y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } \tan x = \sin x \quad 0 < x < \frac { \pi } { 2 }$$ into the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} x } + 2 w \operatorname { cosec } 2 x = \sec x \quad 0 < x < \frac { \pi } { 2 }$$ (c) By solving differential equation (II), determine a general solution of differential equation (I) in the form \(y ^ { 2 } = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a function in terms of \(\cos x\) $$\text { [You may use without proof } \left. \int \operatorname { cosec } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \ln | \tan x | \text { (+ constant) } \right]$$

1. A sample of bacteria in a sealed container is being studied.

The number of bacteria, \(P\), in thousands, is modelled by the differential equation $$( 1 + t ) \frac { \mathrm { d } P } { \mathrm {~d} t } + P = t ^ { \frac { 1 } { 2 } } ( 1 + t )$$ where \(t\) is the time in hours after the start of the study.
Initially, there are exactly 5000 bacteria in the container.

1. Determine, according to the model, the number of bacteria in the container 8 hours after the start of the study.
2. Find, according to the model, the rate of change of the number of bacteria in the container 4 hours after the start of the study.
3. State a limitation of the model.

10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time \(t\) minutes after filling begins the volume of water in the pool is \(V\) litres. The pool has a small leak and loses water at a rate of \(0.01 V\) litres per minute. The differential equation satisfied by \(V\) and \(t\) is of the form \(\frac { \mathrm { d } V } { \mathrm {~d} t } = a - b V\).

1. Write down the values of the constants \(a\) and \(b\).
2. Solve the differential equation and find the value of \(t\) when \(V = 1000\).
3. Obtain an expression for \(V\) in terms of \(t\) and hence state what happens to \(V\) as \(t\) becomes large.

17 In a chemical process, a vessel contains 1 litre of pure water. A liquid chemical is then passed into the top of the vessel at a constant rate of \(a\) litres per minute and thoroughly mixed with the water. At the same time, the resulting mixture is drawn from the bottom of the vessel at a constant rate of \(b\) litres per minute. You may assume that the chemical mixes instantly and uniformly with the water. After \(t\) minutes, the mixture in the vessel contains \(x\) litres of the chemical.

1. 1. Show that the proportion of chemical present in the vessel after \(t\) minutes is $$\frac { x } { 1 + ( a - b ) t } .$$
   2. Hence show that \(\frac { d x } { d t } + \frac { b x } { 1 + ( a - b ) t } = a\).
2. First, consider the case where \(\mathbf { b } = \mathbf { a }\).
   1. Solve the differential equation to find \(x\) in terms of \(a\) and \(t\).
   2. Given that after 1 minute the vessel contains equal amounts of water and chemical, find the rate of inflow of chemical.
3. Now consider the case where \(\mathrm { b } = 2 \mathrm { a }\).
   1. Explain why the differential equation in part (a)(ii) is now invalid for \(\mathrm { t } \geqslant \frac { 1 } { \mathrm { a } }\).
   2. Find the maximum amount of chemical in the vessel.

Solve the differential equation \(\frac{dy}{dx} - 3y = x\) to obtain \(y\) as a function of \(x\). (Total 5 marks)

Solve the differential equation \(\frac{dy}{dx} - 3y = x\) to obtain \(y\) as a function of \(x\). (Total 5 marks)

2

1. Find the values of the constants \(a , b , c\) and \(d\) for which \(a + b x + c \sin x + d \cos x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 10 \sin x - 3 x$$ (4 marks)
2. Hence find the general solution of this differential equation.

3

1. Show that \(\sin x\) is an integrating factor for the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \cot x ) y = 2 \cos x$$
2. Solve this differential equation, given that \(y = 2\) when \(x = \frac { \pi } { 2 }\).

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y \left( 1 + \frac { 3 } { x } \right) = \frac { 1 } { x ^ { 2 } } , \quad x > 0$$

1. Verify that \(x ^ { 3 } \mathrm { e } ^ { x }\) is an integrating factor for the differential equation.
2. Find the general solution of the differential equation.
3. Given that \(y = 1\) at \(x = 1\), find \(y\) at \(x = 2\).
   (3)(Total 10 marks)

7

1. Show that an appropriate integrating factor for $$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$ is \(x + \sqrt { x ^ { 2 } - 1 }\).
2. Hence find the solution of the differential equation $$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$ for which \(y = 1\) when \(x = \frac { 5 } { 4 }\). Give your answer in the form \(y = f ( x )\).

A curve passes through the point \((-2, 4.73)\) and satisfies the differential equation $$\frac{dy}{dx} = \frac{y^2 - x^2}{2x + 3y}$$ Use Euler's step by step method once, and then the midpoint formula $$y_{r+1} = y_{r-1} + 2hf(x_r, y_r), \quad x_{r+1} = x_r + h$$ once, each with a step length of \(0.02\), to estimate the value of \(y\) when \(x = -1.96\) Give your answer to five significant figures. [4 marks]

9. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \mathrm { e } ^ { x ^ { 2 } } .$$ It is given that \(y = 0.2\) at \(x = 0\).

1. Use the approximation \(\frac { y _ { 1 } - y _ { 0 } } { h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 0 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.1\).
2. Use your answer to part (a) and the approximation \(\frac { y _ { 2 } - y _ { 0 } } { 2 h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 1 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.2\). Gives your answer to 4 decimal places.
   (Total 5 marks)

2. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y ^ { 3 } = 4$$

1. Show that $$x \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = a y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + \left( b y ^ { 2 } + c \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are integers to be determined. Given that \(y = 1\) at \(x = 2\)
2. determine the Taylor series expansion for \(y\) in ascending powers of \(( x - 2 )\), up to and including the term in \(( x - 2 ) ^ { 3 }\), giving each coefficient in simplest form.

2 Use the substitution \(u = y ^ { 2 }\) to find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = \frac { \mathrm { e } ^ { x } } { y }$$ for \(y\) in terms of \(x\).

5 The variables \(x\) and \(y\) are related by the differential equation $$x ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x y + x + 1 .$$

1. Use the substitution \(y = u - \frac { 1 } { x }\), where \(u\) is a function of \(x\), to show that the differential equation may be written as $$x ^ { 2 } \frac { \mathrm {~d} u } { \mathrm {~d} x } = u .$$
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = \mathrm { f } ( x )\).

4.The function \(\mathrm { h } ( x )\) has domain \(\mathbb { R }\) and range \(\mathrm { h } ( x ) > 0\) ,and satisfies $$\sqrt { \int \mathrm { h } ( x ) \mathrm { d } x } = \int \sqrt { \mathrm { h } ( x ) } \mathrm { d } x$$

1. By substituting \(\mathrm { h } ( x ) = \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 }\) ,show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( y + c ) ,$$ where \(c\) is constant.
2. Hence find a general expression for \(y\) in terms of \(x\) .
3. Given that \(\mathrm { h } ( 0 ) = 1\) ,find \(\mathrm { h } ( x )\) .

8

1. Using partial fractions, find $$\int \frac { 1 } { y ( 4 - y ) } \mathrm { d } y$$
2. Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 4 - y ) ,$$ obtaining an expression for \(y\) in terms of \(x\).
3. State what happens to the value of \(y\) if \(x\) becomes very large and positive.

8

1. Using partial fractions, find $$\int \frac { 1 } { y ( 4 - y ) } \mathrm { d } y$$
2. Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 4 - y ) ,$$ obtaining an expression for \(y\) in terms of \(x\).
3. State what happens to the value of \(y\) if \(x\) becomes very large and positive.

1. (a) Express \(\frac { 1 } { x ( 2 x - 1 ) }\) in partial fractions.

The height above ground, \(h\) metres, of a carriage on a fairground ride is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 1 } { 50 } h ( 2 h - 1 ) \cos \left( \frac { t } { 10 } \right)$$ where \(t\) seconds is the time after the start of the ride.
Given that, at the start of the ride, the carriage is 2.5 m above ground,
(b) solve the differential equation to show that, according to the model, $$h = \frac { 5 } { 10 - 8 \mathrm { e } ^ { k \sin \left( \frac { t } { 10 } \right) } }$$ where \(k\) is a constant to be found.
(c) Hence find, according to the model, the time taken for the carriage to reach its maximum height above ground for the 3rd time.
Give your answer to the nearest second.
(Solutions relying entirely on calculator technology are not acceptable.)

1. Use the substitution \(y = xz\) to find the general solution of the differential equation $$x \frac{dy}{dx} - y = x \cos \left(\frac{y}{x}\right),$$ giving your answer in a form without logarithms. (You may quote an appropriate result given in the List of Formulae (MF1).) [6]
2. Find the solution of the differential equation for which \(y = \pi\) when \(x = 4\). [2]

1 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ^ { 2 } + y ^ { 2 } } { x y } .$$

1. Use the substitution \(y = u x\), where \(u\) is a function of \(x\), to obtain the differential equation $$x \frac { \mathrm {~d} u } { \mathrm {~d} x } = \frac { 2 } { u } .$$
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. Show that the substitution \(y = vx\) transforms the differential equation $$3xy^2 \frac{dv}{dx} = v^4 + y^3$$ [I] into the differential equation $$3x v^2 \frac{dv}{dx} = 1 - 2v^3$$ [II] [3]
2. By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = f(x)\). [6]

Given that \(y = 2\) at \(x = 1\),

1. find the value of \(\frac{dy}{dx}\) at \(x = 1\). [2]

7 A curve is such that the gradient at a general point with coordinates \(( x , y )\) is proportional to \(\frac { y } { \sqrt { x + 1 } }\). The curve passes through the points with coordinates \(( 0,1 )\) and \(( 3 , e )\). By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).

10 A model for the height, \(h\) metres, of a certain type of tree at time \(t\) years after being planted assumes that, while the tree is growing, the rate of increase in height is proportional to \(( 9 - h ) ^ { \frac { 1 } { 3 } }\). It is given that, when \(t = 0 , h = 1\) and \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.2\).

1. Show that \(h\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.1 ( 9 - h ) ^ { \frac { 1 } { 3 } } .$$
2. Solve this differential equation, and obtain an expression for \(h\) in terms of \(t\).
3. Find the maximum height of the tree and the time taken to reach this height after planting.
4. Calculate the time taken to reach half the maximum height.

9. \includegraphics[max width=\textwidth, alt={}, center]{ab2949b2-11f2-4682-ab0c-25ecee2d665a-4_268_648_1169_623} Two tanks, \(A\) and \(B\), each have a capacity of 800 litres. At time \(t = 0\) both tanks are full of pure water. When \(t > 0\), water flows in the following ways:

• Water with a salt concentration of \(\mu\) grams per litre flows into tank \(A\) at a constant rate
• Water flows from tank \(A\) to tank \(B\) at a rate of 16 litres per minute
• Water flows from tank \(B\) to tank \(A\) at a rate of \(r\) litres per minute
• Water flows out of tank \(B\) through a waste pipe
• The amount of water in each tank remains at 800 litres.

This system is represented by the coupled differential equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 36 - 0.02 x + 0.005 y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.02 x - 0.02 y \end{aligned}$$ Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\).

1. Using the substitution \(t = x^2\), or otherwise, find $$\int x^3 e^{-x^2} \, dx.$$ [6]
2. Find the general solution of the differential equation $$x\frac{dy}{dx} + 3y = xe^{-x^2}, \quad x > 0.$$ [4]

6. (a) Using the substitution \(t = x ^ { 2 }\), or otherwise, find $$\int x ^ { 3 } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x$$ (b) Find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x \mathrm { e } ^ { - x ^ { 2 } } , \quad x > 0$$

7

1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \frac { x } { 2 } \sqrt { x ^ { 2 } - 9 } - \frac { 9 } { 2 } \cosh ^ { - 1 } \frac { x } { 3 } \right) = \sqrt { x ^ { 2 } - 9 }$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_67_1579_413_324} \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_77_1581_497_322}
2. Find the solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = x ^ { 2 } \sqrt { x ^ { 2 } - 9 }$$ given that \(y = 1\) when \(x = 3\). Give your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-14_2716_35_143_2012}

1. Solve the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - 2 \mathbf { r } = \mathbf { 0 }$$ given that when \(t = 0 , \mathbf { r } . \mathbf { j } = 0\) and \(\mathbf { r } \times \mathbf { j } = \mathbf { i } + \mathbf { k }\).

3. The position vector \(\mathbf { r }\) of a particle \(P\) at time \(t\) satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 4 \mathbf { i }$$ Given that the position vector of \(P\) at time \(t = 0\) is \(2 \mathbf { j }\), find the position vector of \(P\) at time \(t\).
(Total 6 marks)

3. A particle \(P\) moves in the \(x y\)-plane in such a way that its position vector \(\mathbf { r }\) metres at time \(t\) seconds, where \(0 \leqslant t < \pi\), satisfies the differential equation $$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$ When \(t = 0\), the particle is at the point with position vector \(( - \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).

7

1. Express \(\frac { 1 } { x } + \frac { 1 } { A - x }\) as a single fraction. The population of fish in a lake is modelled by the differential equation \(\frac { d x } { d t } = \frac { x ( 400 - x ) } { 400 }\) where \(x\) is the number of fish and \(t\) is the time in years.
   When \(t = 0 , x = 100\).
2. In this question you must show detailed reasoning. Find the number of fish in the lake when \(t = 10\), as predicted by the model.

9 In an experiment, at time \(t\) minutes there is \(Q\) grams of substance present.
It is known that the substance decays at a rate that is proportional to \(1 + Q ^ { 2 }\). Initially there are 100 grams of the substance present and after 100 minutes there are 50 grams present. Find the amount of the substance present after 400 minutes.

5 In a certain chemical process a substance \(A\) reacts with another substance \(B\). The masses in grams of \(A\) and \(B\) present at time \(t\) seconds after the start of the process are \(x\) and \(y\) respectively. It is given that \(\frac { \mathrm { d } y } { \mathrm {~d} t } = - 0.6 x y\) and \(x = 5 \mathrm { e } ^ { - 3 t }\). When \(t = 0 , y = 70\).

1. Form a differential equation in \(y\) and \(t\). Solve this differential equation and obtain an expression for \(y\) in terms of \(t\).
2. The percentage of the initial mass of \(B\) remaining at time \(t\) is denoted by \(p\). Find the exact value approached by \(p\) as \(t\) becomes large.

Show that the substitution \(y = x z\) reduces the differential equation $$\frac { 1 } { x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { 6 } { x } - \frac { 2 } { x ^ { 2 } } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( \frac { 9 } { x } - \frac { 6 } { x ^ { 2 } } + \frac { 2 } { x ^ { 3 } } \right) y = 169 \sin 2 x$$ to the differential equation $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 9 z = 169 \sin 2 x$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , z = - 10\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = 5\).

9 The temperature of a quantity of liquid at time \(t\) is \(\theta\). The liquid is cooling in an atmosphere whose temperature is constant and equal to \(A\). The rate of decrease of \(\theta\) is proportional to the temperature difference \(( \theta - A )\). Thus \(\theta\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - A )$$ where \(k\) is a positive constant.

1. Find, in any form, the solution of this differential equation, given that \(\theta = 4 A\) when \(t = 0\).
2. Given also that \(\theta = 3 A\) when \(t = 1\), show that \(k = \ln \frac { 3 } { 2 }\).
3. Find \(\theta\) in terms of \(A\) when \(t = 2\), expressing your answer in its simplest form.

1. Use the substitution \(z = x + y\) to show that the differential equation $$\frac{dy}{dx} = \frac{x + y + 3}{x + y - 1} \qquad (A)$$ may be written in the form \(\frac{dz}{dx} = \frac{2(z + 1)}{z - 1}\). [3]
2. Hence find the general solution of the differential equation (A). [4]

1. Use the substitution \(z = x + y\) to show that the differential equation $$\frac{dy}{dx} = \frac{x + y + 3}{x + y - 1} \qquad (A)$$ may be written in the form \(\frac{dz}{dx} = \frac{2(z + 1)}{z - 1}\). [3]
2. Hence find the general solution of the differential equation (A). [4]