4.10d Second order homogeneous: auxiliary equation method

5

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$
2. It is given that \(y = \mathrm { f } ( x )\) is the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$ such that \(\mathrm { f } ( 0 ) = 0\) and \(\mathrm { f } ^ { \prime } ( 0 ) = 0\).
   1. Show that \(f ^ { \prime \prime } ( 0 ) = 0\).
   2. Find the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\mathrm { f } ( x )\).
      [0pt] [3 marks]

6 A differential equation is given by $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$

1. Show that the substitution \(x = \mathrm { e } ^ { 2 t }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = 4 t ^ { 2 } + 5 \mathrm { e } ^ { - t }$$
2. Hence find the general solution of the differential equation $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$
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7 Find the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 10 \mathrm { e } ^ { 4 x } + 8 \sin 2 x + 4 \cos 2 x$$ given that \(y = 2.5\) when \(x = 0\) and \(y = \frac { \pi } { 4 }\) when \(x = \frac { \pi } { 4 }\).
[0pt] [10 marks]

5

1. Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = 0\).
2. Hence find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = x ( 4 - 5 x )\).

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

11 A company is required to weigh any goods before exporting them overseas. When a crate is placed on a set of weighing scales, the mass displayed takes time to settle down to its final value. The company wishes to model the mass, \(m \mathrm {~kg}\), which is displayed \(t\) seconds after a crate X is placed on the scales.
For the displayed mass it is assumed that the rate of change of the quantity \(\left( 0.5 \frac { \mathrm {~d} m } { \mathrm {~d} t } + m \right)\) with respect to time is proportional to \(( 80 - m )\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 2 \mathrm {~km} = 160 \mathrm { k }\), where \(k\) is a real constant. It is given that the complementary function for the differential equation in part (i) is \(\mathrm { e } ^ { \lambda t } ( A \cos 2 t + B \sin 2 t )\), where \(A\) and \(B\) are arbitrary constants.
2. Show that \(k = \frac { 5 } { 2 }\) and state the value of the constant \(\lambda\). When X is initially placed on the scales the displayed mass is zero and the rate of increase of the displayed mass is \(160 \mathrm {~kg} \mathrm {~s} ^ { - 1 }\).
3. Find \(m\) in terms of \(t\).
4. Describe the long term behaviour of \(m\).
5. With reference to your answer to part (iv), comment on a limitation of the model.
6. (a) Find the value of \(m\) that corresponds to the stationary point on the curve \(m = \mathrm { f } ( t )\) with the smallest positive value of \(t\).
   (b) Interpret this value of \(m\) in the context of the model.
7. Adapt the differential equation \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 5 m = 400\) to model the mass displayed \(t\) seconds after a crate Y , of mass 100 kg , is placed on the scales. \section*{END OF QUESTION PAPER} \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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5 A capacitor is an electrical component which stores charge. The value of the charge stored by the capacitor, in suitable units, is denoted by \(Q\). The capacitor is placed in an electrical circuit. At any time \(t\) seconds, where \(t \geqslant 0 , Q\) can be modelled by the differential equation \(\frac { d ^ { 2 } Q } { d t ^ { 2 } } - 2 \frac { d Q } { d t } - 15 Q = 0\). Initially the charge is 100 units and it is given that \(Q\) tends to a finite limit as \(t\) tends to infinity.

1. Determine the charge on the capacitor when \(t = 0.5\).
2. Determine the finite limit of \(Q\) as \(t\) tends to infinity.

5. A light elastic string, of natural length \(2 a\) and modulus of elasticity \(m g\), has a particle \(P\) of mass \(m\) attached to its mid-point. One end of the string is attached to a fixed point \(A\) and the other end is attached to a fixed point \(B\) which is at a distance \(4 a\) vertically below \(A\).

1. Show that \(P\) hangs in equilibrium at the point \(E\) where \(A E = \frac { 5 } { 2 } a\). The particle \(P\) is held at a distance \(3 a\) vertically below \(A\) and is released from rest at time \(t = 0\). When the speed of the particle is \(v\), there is a resistance to motion of magnitude \(2 m k v\), where \(k = \sqrt { } \left( \frac { g } { a } \right)\). At time \(t\) the particle is at a distance \(\left( \frac { 5 } { 2 } a + x \right)\) from \(A\).
2. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = 0$$
3. Hence find \(x\) in terms of \(t\).

5. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2 m a k ^ { 2 }\), where \(k\) is a positive constant. The other end of the string is attached to a fixed point \(A\). At time \(t = 0 , P\) is released from rest from a point which is a distance \(2 a\) vertically below \(A\). When \(P\) is moving with speed \(v\), the air resistance has magnitude \(2 m k v\). At time \(t\), the extension of the string is \(x\).

1. Show that, while the string is taut, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = g$$ You are given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - k t } ( C \sin k t + D \cos k t ) + \frac { g } { 2 k ^ { 2 } } , \quad \text { where } C \text { and } D \text { are constants. }$$
2. Find the value of \(C\) and the value of \(D\). Assuming that the string remains taut,
3. find the value of \(t\) when \(P\) first comes to rest,
4. show that \(2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)\).

7. A particle \(P\) of mass 0.5 kg is attached to the end \(A\) of a light elastic spring \(A B\), of natural length 0.6 m and modulus of elasticity 2.7 N . At time \(t = 0\) the end \(B\) of the spring is held at rest and \(P\) hangs at rest at the point \(C\) which is vertically below \(B\). The end \(B\) is then moved along the line of the spring so that, at time \(t\) seconds, the downwards displacement of \(B\) from its initial position is \(4 \sin 2 t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of \(P\) below \(C\) is \(y\) metres.

1. Show that $$y + \frac { 49 } { 45 } = x + 4 \sin 2 t$$
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 36 \sin 2 t$$ Given that \(y = \frac { 36 } { 5 } \sin 2 t\) is a particular integral of this differential equation,
3. find \(y\) in terms of \(t\),
4. find the speed of \(P\) when \(t = \frac { 1 } { 3 } \pi\).

4. A particle \(P\) of mass 9 kg moves along the horizontal positive \(x\)-axis under the action of a force directed towards the origin. At time \(t\) seconds, the displacement of \(P\) from \(O\) is \(x\) metres, \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the force has magnitude \(16 x\) newtons. The particle \(P\) is also subject to a resistive force of magnitude \(24 v\) newtons.

1. Show that the equation of motion of \(P\) is $$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 24 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 16 x = 0$$ It is given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - \frac { 4 } { 3 } t } ( A t + B )$$ where \(A\) and \(B\) are arbitrary constants.
   When \(t = \frac { 3 } { 4 } , P\) is travelling towards \(O\) with its maximum speed of \(8 \mathrm { e } ^ { - 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(x = d\).
2. Find the value of \(d\).
3. Find the value of \(x\) when \(t = 0\)

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 \(t = \frac { 1 } { 2 } \ln 2\).

6. A particle \(P\) of mass 2 kg moves in the \(x - y\) plane. At time \(t\) seconds its position vector is \(\mathbf { r }\) metres. When \(t = 0\), the position vector of \(P\) is \(\mathbf { i }\) metres and the velocity of \(P\) is ( \(- \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The vector \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = \mathbf { 0 }$$

1. Find \(\mathbf { r }\) in terms of \(t\).
2. Show that the speed of \(P\) at time \(t\) is \(\mathrm { e } ^ { - t } \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Find, in terms of e, the loss of kinetic energy of \(P\) in the interval \(t = 0\) to \(t = 1\).

2. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$ When \(t = 0\), the velocity of \(P\) is \(( 8 \mathbf { i } - 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
Find the velocity of \(P\) when \(t = \frac { 2 } { 3 } \ln 2\).
(7)

2. The velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$ The position vector of \(P\) at time \(t\) seconds is \(\mathbf { r }\) metres.
Given that at \(t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )\) and \(\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )\), find \(\mathbf { r }\) at time \(t\) seconds.

1. At time \(t = 0\), the position vector of a particle \(P\) is \(- 3 \mathbf { j } \mathrm {~m}\). At time \(t\) seconds, the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that

$$\mathbf { v } - 2 \mathbf { r } = 4 \mathrm { e } ^ { t } \mathbf { j }$$ find the time when \(P\) passes through the origin.

1. A particle \(P\) is moving in a plane. At time \(t\) seconds the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). When \(t = \frac { \pi } { 2 } , P\) is instantaneously at rest at the point with position vector \(( \mathbf { i } - \mathbf { j } ) \mathrm { m }\).

Given that \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = ( 3 \sin t ) \mathbf { i }$$ find \(\mathbf { v }\) in terms of \(t\).
(13)

3. At time \(t\) seconds, the position vector \(\mathbf { r }\) metres of a particle \(P\), relative to a fixed origin \(O\), satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 3 \mathbf { r } = \mathbf { 0 }$$ At time \(t = 0 , P\) is at the point with position vector \(2 \mathbf { i } \mathrm {~m}\) and is moving with velocity \(2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Find the position vector of \(P\) when \(t = \ln 2\).
(10 marks)

1. A scientist is studying the effect of introducing a population of white-clawed crayfish into a population of signal crayfish.
   At time \(t\) years, the number of white-clawed crayfish, \(w\), and the number of signal crayfish, \(s\), are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } w } { \mathrm {~d} t } = \frac { 5 } { 2 } ( w - s ) \\ & \frac { \mathrm { d } s } { \mathrm {~d} t } = \frac { 2 } { 5 } w - 90 \mathrm { e } ^ { - t } \end{aligned}$$

1. Show that $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} w } { \mathrm {~d} t } + 2 w = 450 \mathrm { e } ^ { - t }$$
2. Find a general solution for the number of white-clawed crayfish at time \(t\) years.
3. Find a general solution for the number of signal crayfish at time \(t\) years. The model predicts that, at time \(T\) years, the population of white-clawed crayfish will have died out. Given that \(w = 65\) and \(s = 85\) when \(t = 0\)
4. find the value of \(T\), giving your answer to 3 decimal places.
5. Suggest a limitation of the model.

1. Two compounds, \(X\) and \(Y\), are involved in a chemical reaction. The amounts in grams of these compounds, \(t\) minutes after the reaction starts, are \(x\) and \(y\) respectively and are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = - 5 x + 10 y - 30 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = - 2 x + 3 y - 4 \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 50$$
2. Find, according to the model, a general solution for the amount in grams of compound \(X\) present at time \(t\) minutes.
3. Find, according to the model, a general solution for the amount in grams of compound \(Y\) present at time \(t\) minutes. Given that \(x = 2\) and \(y = 5\) when \(t = 0\)
4. find
   1. the particular solution for \(x\),
   2. the particular solution for \(y\). A scientist thinks that the chemical reaction will have stopped after 8 minutes.
5. Explain whether this is supported by the model.

1. A company plans to build a new fairground ride. The ride will consist of a capsule that will hold the passengers and the capsule will be attached to a tall tower. The capsule is to be released from rest from a point half way up the tower and then made to oscillate in a vertical line.

The vertical displacement, \(x\) metres, of the top of the capsule below its initial position at time \(t\) seconds is modelled by the differential equation, $$m \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 200 \cos t , \quad t \geqslant 0$$ where \(m\) is the mass of the capsule including its passengers, in thousands of kilograms.
The maximum permissible weight for the capsule, including its passengers, is 30000 N .
Taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\) and assuming the capsule is at its maximum permissible weight,

1. 1. explain why the value of \(m\) is 3
   2. show that a particular solution to the differential equation is $$x = 40 \sin t - 20 \cos t$$
   3. hence find the general solution of the differential equation.
2. Using the model, find, to the nearest metre, the vertical distance of the top of the capsule from its initial position, 9 seconds after it is released.

1. An engineer is investigating the motion of a sprung diving board at a swimming pool.

Let \(E\) be the position of the end of the diving board when it is at rest in its equilibrium position and when there is no diver standing on the diving board.
A diver jumps from the diving board.
The vertical displacement, \(h \mathrm {~cm}\), of the end of the diving board above \(E\) is modelled by the differential equation $$4 \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} h } { \mathrm {~d} t } + 37 h = 0$$ where \(t\) seconds is the time after the diver jumps.

1. Find a general solution of the differential equation. When \(t = 0\), the end of the diving board is 20 cm below \(E\) and is moving upwards with a speed of \(55 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).
2. Find, according to the model, the maximum vertical displacement of the end of the diving board above \(E\).
3. Comment on the suitability of the model for large values of \(t\).

1. A scientist is investigating the concentration of antibodies in the bloodstream of a patient following a vaccination.
   The concentration of antibodies, \(x\), measured in micrograms ( \(\mu \mathrm { g }\) ) per millilitre ( ml ) of blood, is modelled by the differential equation

$$100 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 60 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 26$$ where \(t\) is the number of weeks since the vaccination was given.

1. Find a general solution of the differential equation. Initially,
   • there are no antibodies in the bloodstream of the patient
   • the concentration of antibodies is estimated to be increasing at \(10 \mu \mathrm {~g} / \mathrm { ml }\) per week
   • Find, according to the model, the maximum concentration of antibodies in the bloodstream of the patient after the vaccination.
   A second dose of the vaccine has to be given to try to ensure that it is fully effective. It is only safe to give the second dose if the concentration of antibodies in the bloodstream of the patient is less than \(5 \mu \mathrm {~g} / \mathrm { ml }\).
2. Determine whether, according to the model, it is safe to give the second dose of the vaccine to the patient exactly 10 weeks after the first dose.

1. At the start of the year 2000, a survey began of the number of foxes and rabbits on an island.

At time \(t\) years after the survey began, the number of foxes, \(f\), and the number of rabbits, \(r\), on the island are modelled by the differential equations $$\begin{aligned} & \frac { \mathrm { d } f } { \mathrm {~d} t } = 0.2 f + 0.1 r \\ & \frac { \mathrm {~d} r } { \mathrm {~d} t } = - 0.2 f + 0.4 r \end{aligned}$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } f } { \mathrm {~d} t ^ { 2 } } - 0.6 \frac { \mathrm {~d} f } { \mathrm {~d} t } + 0.1 f = 0\)
2. Find a general solution for the number of foxes on the island at time \(t\) years.
3. Hence find a general solution for the number of rabbits on the island at time \(t\) years. At the start of the year 2000 there were 6 foxes and 20 rabbits on the island.
4. 1. According to this model, in which year are the rabbits predicted to die out?
   2. According to this model, how many foxes will be on the island when the rabbits die out?
   3. Use your answers to parts (i) and (ii) to comment on the model.