4.10c Integrating factor: first order equations

6

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
   [0pt] [6 marks]
2. Show that the substitution \(u = x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\) transforms the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
3. Hence, given that \(x > 0\), find the general solution of the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ [2 marks]

2 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \tan ^ { 3 } x \sec x$$ given that \(y = 2\) when \(x = \frac { \pi } { 3 }\).
[0pt] [9 marks]

5

1. Express \(\frac { 1 } { ( 1 + x ) ( 2 + x ) }\) in the form \(\frac { A } { 1 + x } + \frac { B } { 2 + x }\), where \(A\) and \(B\) are integers.
2. Use the substitution \(u = \frac { \mathrm { d } y } { \mathrm {~d} x }\) to solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { 1 } { ( 1 + x ) ( 2 + x ) } \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + x } { 1 + x }$$ given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).
   [0pt] [11 marks]

10 A particle \(B\), of mass 3 kg , moves in a straight line and has velocity \(v \mathrm {~ms} ^ { - 1 }\).
At time \(t\) seconds, where \(0 \leqslant t < \frac { 1 } { 4 } \pi\), a variable force of \(- ( 15 \sin 4 \mathrm { t } + 6 \mathrm { v } \tan 2 \mathrm { t } )\) Newtons is applied to \(B\). There are no other forces acting on \(B\). Initially, when \(t = 0 , B\) has velocity \(4.5 \mathrm {~ms} ^ { - 1 }\). The motion of \(B\) can be modelled by the differential equation \(\frac { d v } { d t } + P ( t ) v = Q ( t )\) where \(P ( t )\) and \(\mathrm { Q } ( \mathrm { t } )\) are functions of \(t\).

1. Find the functions \(\mathrm { P } ( \mathrm { t } )\) and \(\mathrm { Q } ( \mathrm { t } )\).
2. Using an integrating factor, determine the first time at which \(B\) is stationary according to the model.

10 A particle of mass 0.5 kg is initially at point \(O\). It moves from rest along the \(x\)-axis under the influence of two forces \(F _ { 1 } \mathrm {~N}\) and \(F _ { 2 } \mathrm {~N}\) which act parallel to the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\). \(F _ { 1 }\) is acting in the direction of motion of the particle and \(F _ { 2 }\) is resisting motion.
In an initial model

• \(F _ { 1 }\) is proportional to \(t\) with constant of proportionality \(\lambda > 0\),
• \(F _ { 2 }\) is proportional to \(v\) with constant of proportionality \(\mu > 0\).
  1. Show that the motion of the particle can be modelled by the following differential equation.

$$\frac { 1 } { 2 } \frac { d v } { d t } = \lambda t - \mu v$$

Solve the differential equation in part (a), giving the particular solution for \(v\) in terms of \(t\), \(\lambda\) and \(\mu\). You are now given that \(\lambda = 2\) and \(\mu = 1\).

Find a formula for an approximation for \(v\) in terms of \(t\) when \(t\) is large. In a refined model

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.

8 A particle \(P\) of mass 2 kg can only move along the straight line segment \(O A\), where \(O A\) is on a rough horizontal surface. The particle is initially at rest at \(O\) and the distance \(O A\) is 0.9 m . When the time is \(t\) seconds the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). \(P\) is subject to a force of magnitude \(4 \mathrm { e } ^ { - 2 t } \mathrm {~N}\) in the direction of \(A\) for any \(t \geqslant 0\). The resistance to the motion of \(P\) is modelled as being proportional to \(v\). At the instant when \(t = \ln 2 , v = 0.5\) and the resultant force on \(P\) is 0 N .

1. Show that, according to the model, \(\frac { d v } { d t } + v = 2 e ^ { - 2 t }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. By considering the behaviour of \(v\) as \(t\) becomes large explain why, according to the model, \(P\) 's speed must reach a maximum value for some \(t > 0\).
4. Determine the maximum speed considered in part (c).
5. Determine the greatest value of \(t\) for which the model is valid.

5 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, \(\frac { d v } { d t } = \frac { 10 f - 4 v } { 45 }\). \item

1. By solving the differential equation in part (a), show that \(\mathrm { v } = \frac { 1 } { 2 } \left( 5 \mathrm { f } - ( 5 \mathrm { f } - 2 \mathrm { u } ) \mathrm { e } ^ { - \frac { 4 } { 45 } \mathrm { t } } \right)\).
2. Describe briefly how, according to the model, the speed of \(P\) varies over time in each of the following cases.

8. Four uniform rods, each of mass \(m\) and length \(2 a\), are joined together at their ends to form a plane rigid square framework \(A B C D\) of side \(2 a\). The framework is free to rotate in a vertical plane about a fixed smooth horizontal axis through \(A\). The axis is perpendicular to the plane of the framework.

1. Show that the moment of inertia of the framework about the axis is \(\frac { 40 m a ^ { 2 } } { 3 }\). The framework is slightly disturbed from rest when \(C\) is vertically above \(A\). Find
2. the angular acceleration of the framework when \(A C\) is horizontal,
3. the angular speed of the framework when \(A C\) is horizontal,
4. the magnitude of the force acting on the framework at \(A\) when \(A C\) is horizontal.

3. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf { r }\) metres, relative to a fixed origin. The particle moves in such a way that $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$ At \(t = 0 , P\) is moving with velocity ( \(8 \mathbf { i } - 6 \mathbf { j }\) ) \(\mathrm { m } \mathrm { s } ^ { - 1 }\).
Find the speed of \(P\) when \(t = \frac { 1 } { 2 } \ln 2\).

2. With respect to a fixed origin \(O\), the position vector, \(\mathbf { r }\) metres, of a particle \(P\) at time \(t\) seconds satisfies $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \mathbf { r } = ( \mathbf { i } - \mathbf { j } ) \mathrm { e } ^ { - 2 t } .$$ Given that \(P\) is at \(O\) when \(t = 0\), find

1. \(\mathbf { r }\) in terms of \(t\),
2. a cartesian equation of the path of \(P\).

2. At time \(t\) seconds the position vector of a particle \(P\), relative to a fixed origin \(O\), is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 3 \mathrm { e } ^ { - t } \mathbf { j }$$ Given that \(\mathbf { r } = 2 \mathbf { i } - \mathbf { j }\) when \(t = 0\), find \(\mathbf { r }\) in terms of \(t\).
(Total 7 marks)

1. A particle moves in a plane so that its position vector \(\mathbf { r }\) metres at time \(t\) seconds satisfies the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + ( \tan t ) \mathbf { r } = \left( \cos ^ { 2 } t \right) \mathbf { i } - ( 3 \cos t ) \mathbf { j } , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ When \(t = 0\), the particle is at the point with position vector \(4 \mathbf { j } \mathrm {~m}\). Find \(\mathbf { r }\) in terms of \(t\).

2. A particle \(P\) moves so that its position vector, \(\mathbf { r }\) metres, at time \(t\) seconds, where \(0 \leqslant t < \frac { \pi } { 2 }\), satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - ( \tan t ) \mathbf { r } = ( \sin t ) \mathbf { i }$$ When \(t = 0 , \mathbf { r } = - \frac { 1 } { 2 } \mathbf { i }\).
Find \(\mathbf { r }\) in terms of \(t\).

2. [In this question, \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors in a horizontal plane and \(\mathbf { k }\) is a unit vector vertically upwards.] A particle of mass 2 kg moves under the action of a constant gravitational force \(- 19.6 \mathbf { k } \mathrm {~N}\). The particle is subject to a resistive force \(- \mathbf { v }\) newtons, where \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is the velocity of the particle at time \(t\) seconds.

1. By writing down an equation of motion of the particle, show that \(\mathbf { v }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 0.5 \mathbf { v } = - 9.8 \mathbf { k }$$ When \(t = 0 , \mathbf { v } = ( 4 \mathbf { i } - 6 \mathbf { j } + 11.6 \mathbf { k } )\)
2. Find \(\mathbf { v }\) when \(t = \ln 4\)

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 \(t\).

17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is \(m \mathrm {~kg}\), and at time \(t\) seconds, for \(0 \leqslant t \leqslant 10\), the cyclist's velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistance to motion, modelled by a force of magnitude 0.1 mvN , acts on the cyclist during the whole 10 seconds.

1. Explain why modelling the resistance to motion in this way is likely to be more realistic than assuming this force is constant. During the braking phase of the motion, for \(5 \leqslant t \leqslant 10\), the brakes apply an additional constant resistance force of magnitude \(2 m \mathrm {~N}\) and the cyclist does not provide any driving force.
2. Show that, for \(5 \leqslant t \leqslant 10 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = - 2\).
3. 1. Solve the differential equation in part (b).
   2. Hence find the velocity of the cyclist when \(t = 5\). During the acceleration phase ( \(0 \leqslant t \leqslant 5\) ), the cyclist applies a driving force of magnitude directly proportional to \(t\).
4. Show that, for \(0 \leqslant t \leqslant 5 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = \lambda t\), where \(\lambda\) is a positive constant.
5. 1. Show by integration that, for \(0 \leqslant t \leqslant 5 , v = 10 \lambda \left( t - 10 + 10 \mathrm { e } ^ { - 0.1 t } \right)\).
   2. Hence find \(\lambda\).
6. Find the total distance, to the nearest metre, travelled by the cyclist during the motion.

12 Solve the differential equation \(\left( 4 - x ^ { 2 } \right) \frac { d y } { d x } - x y = 1\), given that \(y = 1\) when \(x = 0\), giving your answer in the form \(y = \mathrm { f } ( x )\).

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

17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After \(t\) minutes the mass of salt in the container is \(x\) grams, and you are given that \(x = 0\) when \(t = 0\). In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.

1. 1. Show that the concentration of salt in the liquid after \(t\) minutes is \(\frac { \mathrm { X } } { 1000 - 5 \mathrm { t } }\) grams per litre.
   2. Hence show that the mass of salt in the container is given by the differential equation $$\frac { d x } { d t } + \frac { x } { 200 - t } = 10$$
2. Show by integration that \(\mathrm { x } = 10 ( 200 - \mathrm { t } ) \ln \left( \frac { 200 } { 200 - \mathrm { t } } \right)\).
3. 1. Hence determine the mass of salt in the container when half the liquid is drawn off.
   2. Determine also the time at which the mass of salt in the container is greatest.
4. When the process is run, it is found that the concentration of salt over time is higher than predicted by the model. Suggest a reason for this.

16 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 \(A\). \begin{enumerate}[label=(\alph*)] \item One simple model is to assume that the rate of change of population density is directly proportional to \(A - P\).

1. Formulate a differential equation for this model.
2. Verify that \(P = A \left( 1 - \mathrm { e } ^ { - k t } \right)\), where \(k\) is a positive constant, satisfies
   An alternative model uses the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \mathrm { Q } ( t )$$ where \(\mathrm { Q } ( t )\) is a function of \(t\).
3. Find the integrating factor for this differential equation, showing that it can be written in the $$\text { form } \frac { \sqrt { 1 + t ^ { 2 } } } { t } \text {. }$$
4. Suppose that \(\mathrm { Q } ( t ) = 0\). $$\text { (i) Show that } P = \frac { A t } { \sqrt { 1 + t ^ { 2 } } } \text {. }$$ (ii) Find the time predicted by this model for the population density to reach half its longterm value. Give your answer correct to the nearest minute.
5. Now suppose that \(\mathrm { Q } ( t ) = \frac { t \mathrm { e } ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }\). $$\text { Show that } \left. P = \frac { A t - t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } } \text {. [You may assume that } \lim _ { t \rightarrow \infty } t e ^ { - t } = 0 . \right]$$ It is found that the long-term value of \(P\) is 10, and \(P\) reaches half this value after 37 minutes.
6. Determine which of the models proposed in parts (c) and (d) is more consistent with these data.

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.

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 \(P _ { 0 } = 1\).

1. \(\left[ \begin{array} { l } \text { With respect to the right-hand rule, a rotation through } \theta ^ { \circ } \text { anticlockwise about } \\ \text { the } z \text {-axis is represented by the matrix } \\ \qquad \left( \begin{array} { c c c } \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right) \end{array} \right]\)

Given that the matrix \(\mathbf { M }\), where $$\mathbf { M } = \left( \begin{array} { c c c } - \frac { \sqrt { 3 } } { 2 } & \frac { 1 } { 2 } & 0 \\ - \frac { 1 } { 2 } & - \frac { \sqrt { 3 } } { 2 } & 0 \\ 0 & 0 & 1 \end{array} \right)$$ represents a rotation through \(\alpha ^ { \circ }\) anticlockwise about the \(z\)-axis with respect to the right-hand rule,

1. determine the value of \(\alpha\).
2. Hence determine the smallest possible positive integer value of \(k\) for which \(\mathbf { M } ^ { k } = \mathbf { I }\) The \(3 \times 3\) matrix \(\mathbf { N }\) represents a reflection in the plane with equation \(y = 0\)
3. Write down the matrix \(\mathbf { N }\). The point \(A\) has coordinates (-2, 4, 3)
   The point \(B\) is the image of the point \(A\) under the transformation represented by matrix \(\mathbf { M }\) followed by the transformation represented by matrix \(\mathbf { N }\).
4. Show that the coordinates of \(B\) are \(( 2 + \sqrt { 3 } , 2 \sqrt { 3 } - 1,3 )\) Given that \(O\) is the origin,
5. show that, to 3 significant figures, the size of angle \(A O B\) is \(66.9 ^ { \circ }\)
6. Hence determine the area of triangle \(A O B\), giving your answer to 3 significant figures.

1. A tank at a chemical plant has a capacity of 250 litres. The tank initially contains 100 litres of pure water.

Salt water enters the tank at a rate of 3 litres every minute. Each litre of salt water entering the tank contains 1 gram of salt. It is assumed that the salt water mixes instantly with the contents of the tank upon entry.
At the instant when the salt water begins to enter the tank, a valve is opened at the bottom of the tank and the solution in the tank flows out at a rate of 2 litres per minute. Given that there are \(S\) grams of salt in the tank after \(t\) minutes,

1. show that the situation can be modelled by the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } = 3 - \frac { 2 S } { 100 + t }$$
2. Hence find the number of grams of salt in the tank after 10 minutes. When the concentration of salt in the tank reaches 0.9 grams per litre, the valve at the bottom of the tank must be closed.
3. Find, to the nearest minute, when the valve would need to be closed.
4. Evaluate the model.