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 \(x = 0\).

\begin{enumerate} \item Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = x\) to obtain \(y\) as a function of \(x\). \item (a) Simplify the expression \(\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } - ( 3 x - 5 )\), giving your answer in the form \(\frac { a ( x + b ) ( x + c ) } { x - 1 }\), where \(a , b\) and \(c\) are integers.
(b) Hence, or otherwise, solve the inequality \(\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } > 3 x - 5 \quad\) (4)(Total \(\mathbf { 8 }\) marks) \item (a) Find the general solution of the differential equation \(3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = x ^ { 2 }\)
(b) Find the particular solution for which, at \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\).(6)(Total 14 marks) \item The diagram above shows the curve \(C _ { 1 }\) which has polar equation \(\boldsymbol { r } = \boldsymbol { a } ( \mathbf { 3 } + \mathbf { 2 } \boldsymbol { \operatorname { c o s } } \boldsymbol { \theta } ) , 0 \leq \theta < 2 \pi\) and the circle \(C _ { 2 }\) with equation \(\boldsymbol { r } = \mathbf { 4 } \boldsymbol { a } , 0 \leq \theta < 2 \pi\), where \(a\) is a positive constant.

4 Find the solution of the differential equation $$\sin \theta \frac { d y } { d \theta } + y = \tan \frac { 1 } { 2 } \theta$$ where \(0 < \theta < \pi\), given that \(y = 1\) when \(\theta = \frac { 1 } { 2 } \pi\). Give your answer in the form \(y = \mathrm { f } ( \theta )\). [You may use without proof the result that \(\int \operatorname { cosec } \theta d \theta = \ln \tan \frac { 1 } { 2 } \theta\).]

5
2
4 \end{array} \right) + \mu \left( \begin{array} { r } 1
- 2
1 \end{array} \right) \end{aligned}$$ Find, in exact form, the distance between \(l _ { 1 }\) and \(l _ { 2 }\).

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 $$\int \text { uv } \mathrm { d } x = \left( \int \mathrm { u } \mathrm {~d} x \right) \times \left( \int \mathrm { v } \mathrm {~d} x \right) \quad \text { uv } \neq 0$$ （a）show that \(1 = \frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } + \frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) Given also that \(\frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } = \mathrm { sin } ^ { 2 } x\),
（b）use part（a）to write down an expression，in terms of \(x\) ，for \(\frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) ，
（c）show that $$\frac { 1 } { \mathrm { u } } \frac { \mathrm { du } } { \mathrm {~d} x } = \frac { 1 - 2 \sin x \cos x } { \sin ^ { 2 } x }$$ （d）hence use integration to show that \(\mathrm { u } = A \mathrm { e } ^ { - \cot x } \operatorname { cosec } ^ { 2 } x\) ，where \(A\) is an arbitrary constant．
（e）By differentiating \(\mathrm { e } ^ { \tan x }\) find a similar expression for v ．

5. (a) Obtain the general solution of the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } - 0.1 S = t$$ (b) The differential equation in part (a) is used to model the assets, \(\pounds S\) million, of a bank \(t\) years after it was set up. Given that the initial assets of the bank were \(\pounds 200\) million, use your answer to part (a) to estimate, to the nearest \(\pounds\) million, the assets of the bank 10 years after it was set up.

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)

12. The population density \(P\), in suitable units, of a certain bacterium at time \(t\) hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is 5 . The model uses the differential equation $$\frac { d P } { d t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \frac { t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }$$ Find \(P\) as a function of \(t\). [You may assume that \(\lim _ { t \rightarrow \infty } t e ^ { - t } = 0\) ].
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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 $$\sqrt { \int \mathrm { h } ( x ) \mathrm { d } x } = \int \sqrt { \mathrm { h } ( x ) } \mathrm { d } x$$ （a）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．
（b）Hence find a general expression for \(y\) in terms of \(x\) ．
（c）Given that \(\mathrm { h } ( 0 ) = 1\) ，find \(\mathrm { h } ( x )\) ．

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.

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.

10 A balloon in the shape of a sphere has volume \(V\) and radius \(r\). Air is pumped into the balloon at a constant rate of \(40 \pi\) starting when time \(t = 0\) and \(r = 0\). At the same time, air begins to flow out of the balloon at a rate of \(0.8 \pi r\). The balloon remains a sphere at all times.

1. Show that \(r\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { 50 - r } { 5 r ^ { 2 } }$$
2. Find the quotient and remainder when \(5 r ^ { 2 }\) is divided by \(50 - r\).
3. Solve the differential equation in part (a), obtaining an expression for \(t\) in terms of \(r\).
4. Find the value of \(t\) when the radius of the balloon is 12 .

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. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

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. (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 \(\cos x\) $$\text { [You may use without proof } \left. \int \operatorname { cosec } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \ln | \tan x | \text { (+ constant) } \right]$$

4

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

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 \(y ^ { 2 } = \mathrm { f } ( x )\).

6 Use the substitution \(y = v x\) to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), where \(\mathrm { f } ( x )\) is a polynomial in \(x\). [10]
\includegraphics[max width=\textwidth, alt={}, center]{114ece0d-558d-4c02-8a77-034b3681cff9-10_51_1648_527_246}

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.

13. \begin{figure}[h] \end{figure} Figure 4 shows a hemispherical bowl containing some water.
At \(t\) seconds, the height of the water is \(h \mathrm {~cm}\) and the volume of the water is \(V \mathrm {~cm} ^ { 3 }\), where $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 30 - h ) , \quad 0 < h \leqslant 10$$ The water is leaking from a hole in the bottom of the bowl. Given that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - \frac { 1 } { 10 } V\)

1. show that \(\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { h ( 30 - h ) } { 30 ( 20 - h ) }\)
2. Write \(\frac { 30 ( 20 - h ) } { h ( 30 - h ) }\) in partial fraction form. Given that \(h = 10\) when \(t = 0\),
3. use your answers to parts (a) and (b) to find the time taken for the height of the water to fall to 5 cm . Give your answer in seconds to 2 decimal places.

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.

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.

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 $$\left( x ^ { 2 } + 1 \right) \frac { d y } { d x } + y \sqrt { x ^ { 2 } + 1 } = x ^ { 2 } - x \sqrt { x ^ { 2 } + 1 }$$ is \(x + \sqrt { x ^ { 2 } + 1 }\).
2. Hence find the solution of the differential equation $$\left( x ^ { 2 } + 1 \right) \frac { d y } { d x } + y \sqrt { x ^ { 2 } + 1 } = x ^ { 2 } - x \sqrt { x ^ { 2 } + 1 }$$ for which \(y = \ln 2\) when \(x = 0\). Give your answer in the form \(y = f ( x )\).

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. 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.

2 A particle \(Q\) of mass \(m \mathrm {~kg}\) falls from rest under gravity. The motion of \(Q\) is resisted by a force of magnitude \(m k v \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(Q\) at time \(t \mathrm {~s}\) and \(k\) is a positive constant. Find an expression for \(v\) in terms of \(g , k\) and \(t\).

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.

2．The points \(( x , y )\) on the curve \(C\) satisfy $$( x + 1 ) ( x + 2 ) \frac { d y } { d x } = x y$$ The line with equation \(y = 2 x + 5\) is the tangent to \(C\) at a point \(P\) ．
（a）Find the coordinates of \(P\) ．
（b）Find the equation of \(C\) ，giving your answer in the form \(y = \mathrm { f } ( x )\) ．

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.