4.10a General/particular solutions: of differential equations

3 The differential equation $$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y ^ { 3 } = \frac { \cos x } { x }$$ is to be solved for \(x > 0\). Use the substitution \(u = y ^ { 3 }\) to find the general solution for \(y\) in terms of \(x\).

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 \(x\).

7 Find the value of the constant \(\lambda\) such that \(\lambda x \mathrm { e } ^ { - x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 6 \mathrm { e } ^ { - x }$$ Find the solution of the differential equation for which \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) when \(x = 0\).

4 Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 195 \sin 2 t$$

8 Given that $$2 y ^ { 3 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 12 y ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y ^ { 2 } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 17 y ^ { 4 } = 13 \mathrm { e } ^ { - 4 x }$$ and that \(v = y ^ { 4 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 34 v = 26 \mathrm { e } ^ { - 4 x }$$ Hence find the general solution for \(y\) in terms of \(x\).

8

1. Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
   1. \(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the substitution \(x = t ^ { 2 }\), where \(t \geqslant 0\), transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$ (2 marks)
3. Hence find the general solution of the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

7

1. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\);
      (2 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }\).
      (3 marks)
2. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t$$ (2 marks)
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

6 A differential equation is given by $$\left( x ^ { 3 } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 3 x ^ { 2 } \frac { d y } { d x } = 2 - 4 x ^ { 3 }$$

1. Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 x$$ transforms this differential equation into $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } u } { \mathrm {~d} x } = 3 x ^ { 2 } u$$ (4 marks)
2. Hence find the general solution of the differential equation $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 - 4 x ^ { 3 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \(7 \quad\) The curve \(C _ { 1 }\) is defined by \(r = 2 \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\). The curve \(C _ { 2 }\) is defined by \(r = \tan \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\).
   1. Find a cartesian equation of \(C _ { 1 }\).
   2. 1. Prove that the curves \(C _ { 1 }\) and \(C _ { 2 }\) meet at the pole \(O\) and at one other point, \(P\), in the given domain. State the polar coordinates of \(P\).
      2. The point \(A\) is the point on the curve \(C _ { 1 }\) at which \(\theta = \frac { \pi } { 4 }\). The point \(B\) is the point on the curve \(C _ { 2 }\) at which \(\theta = \frac { \pi } { 4 }\). Determine which of the points \(A\) or \(B\) is further away from the pole \(O\), justifying your answer.
      3. Show that the area of the region bounded by the arc \(O P\) of \(C _ { 1 }\) and the arc \(O P\) of \(C _ { 2 }\) is \(a \pi + b \sqrt { 3 }\), where \(a\) and \(b\) are rational numbers.

1

1. Find the values of the constants \(a\) and \(b\) for which \(a x + b\) is a particular integral of the differential equation $$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 5 y = 10 x$$
2. Hence find the general solution of \(2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 5 y = 10 x\).
   [0pt] [3 marks]

3

1. It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = ( 2 x + 1 ) \ln ( x + y )$$ and $$y ( 0 ) = 2$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.1\), to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places.
2. It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) \ln ( x + y )$$ and \(y = 2\) when \(x = 0\).
   1. Use implicit differentiation to find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), giving your answer in terms of \(x\) and \(y\).
   2. Hence find the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(y ( x )\). Give your answer in an exact form.
   3. Use your answer to part (b)(ii) to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places.
      [0pt] [1 mark]

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

4 The resistive force, \(F\), on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, \(r\), the velocity of the sphere, \(v\), and the viscosity of the fluid, \(\eta\). You are given that \(\eta\) is measured in \(\mathrm { N } \mathrm { m } ^ { - 2 } \mathrm {~s}\).

1. By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: \(\mathbf { F } = 6 \pi \eta ^ { \alpha } \mathbf { r } ^ { \beta } \mathbf { v } ^ { \gamma }\).
2. Explain whether or not it is possible to use dimensional analysis to verify that the constant \(6 \pi\) is correct.
3. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\). A sphere of radius \(r\) and mass \(m\) falls vertically from rest through the fluid. After a time \(t\) its velocity is \(v\).
4. By setting up and solving a differential equation, show that \(\mathrm { e } ^ { - \mathrm { kt } } = \frac { \mathrm { g } - \mathrm { kv } } { \mathrm { g } }\) where \(\mathrm { k } = \frac { 6 \pi \eta \mathrm { r } } { \mathrm { m } }\). As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
5. Find, in terms of \(g\) and \(k\), the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
6. Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.

1 A spherical raindrop falls through a stationary cloud. Water condenses on the raindrop and it gains mass at a rate proportional to its surface area. At time \(t\) the radius of the raindrop is \(r\). Initially the raindrop is at rest and \(r = r _ { 0 }\). The density of the water is \(\rho\).

1. Show that \(\frac { \mathrm { d } r } { \mathrm {~d} t } = k\), where \(k\) is a constant. Hence find the mass of the raindrop in terms of \(r _ { 0 } , \rho , k\) and \(t\).
2. Assuming that air resistance is negligible, find the velocity of the raindrop in terms of \(r _ { 0 } , k\) and \(t\).

3 An aeroplane is taking off from a runway. It starts from rest. The resultant force in the direction of motion has power, \(P\) watts, modelled by $$P = 0.0004 m \left( 10000 v + v ^ { 3 } \right) ,$$ where \(m \mathrm {~kg}\) is the mass of the aeroplane and \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity at time \(t\) seconds. The displacement of the aeroplane from its starting point is \(x \mathrm {~m}\). To take off successfully the aeroplane must reach a speed of \(80 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) before it has travelled 900 m .

1. Formulate and solve a differential equation for \(v\) in terms of \(x\). Hence show that the aeroplane takes off successfully.
2. Formulate a differential equation for \(v\) in terms of \(t\). Solve the differential equation to show that \(v = 100 \tan ( 0.04 t )\). What feature of this result casts doubt on the validity of the model?
3. In fact the model is only valid for \(0 \leqslant t \leqslant 11\), after which the power remains constant at the value attained at \(t = 11\). Will the aeroplane take off successfully?

5. A rocket is launched vertically upwards under gravity from rest at time \(t = 0\). The rocket propels itself upward by ejecting burnt fuel vertically downwards at a constant speed \(u\) relative to the rocket. The initial mass of the rocket, including fuel, is \(M\). At time \(t\), before all the fuel has been used up, the mass of the rocket, including fuel, is \(M ( 1 - k t )\) and the speed of the rocket is \(v\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { k u } { 1 - k t } - g\).
2. Hence find the speed of the rocket when \(t = \frac { 1 } { 3 k }\).

6. A rocket-driven car moves along a straight horizontal road. The car has total initial mass \(M\). It propels itself forwards by ejecting mass backwards at a constant rate \(\lambda\) per unit time at a constant speed \(U\) relative to the car. The car starts from rest at time \(t = 0\). At time \(t\) the speed of the car is \(v\). The total resistance to motion is modelled as having magnitude \(k v\), where \(k\) is a constant. Given that \(t < \frac { M } { \lambda }\), show that

1. \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { \lambda U - k v } { M - \lambda t }\),
2. \(v = \frac { \lambda U } { k } \left\{ 1 - \left( 1 - \frac { \lambda t } { M } \right) ^ { \frac { k } { \lambda } } \right\}\).
   (6)
   (Total 13 marks)

7. A motor boat of mass \(M\) is moving in a straight line, with its engine switched off, across a stretch of still water. The boat is moving with speed \(U\) when, at time \(t = 0\), it develops a leak. The water comes in at a constant rate so that at time \(t\), the mass of water in the boat is \(\lambda t\). At time \(t\) the speed of the boat is \(v\) and it experiences a total resistance to motion of magnitude \(2 \lambda v\).

1. Show that \(( M + \lambda t ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 \lambda v = 0\).
   (6)
2. Show that the time taken for the speed of the boat to reduce to \(\frac { 1 } { 2 } U\) is \(\frac { M } { \lambda } \left( 2 ^ { \frac { 1 } { 3 } } - 1 \right)\).
   (6) The boat sinks when the mass of water inside the boat is \(M\).
3. Show that the boat does not sink before the speed of the boat is \(\frac { 1 } { 2 } U\).

4. A particle \(P\), whose initial mass is \(m _ { 0 }\), is projected vertically upwards from the ground at time \(t = 0\) with speed \(\frac { g } { k }\), where \(k\) is a constant. As the particle moves upwards it gains mass by picking up small droplets of moisture from the atmosphere. The droplets are at rest before they are picked up. At time \(t\) the speed of \(P\) is \(v\) and its mass has increased to \(m _ { 0 } \mathrm { e } ^ { k t }\). Assuming that, during the motion, the acceleration due to gravity is constant,

1. show that, while \(P\) is moving upwards, $$k v + \frac { \mathrm { d } v } { \mathrm {~d} t } = - g$$
2. find, in terms of \(m _ { 0 }\), the mass of \(P\) when it reaches its greatest height above the ground.
   (6)

6. A firework rocket, excluding its fuel, has mass \(m _ { 0 } \mathrm {~kg}\). The rocket moves vertically upwards by ejecting burnt fuel vertically downwards with constant speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 } , u > 24.5\), relative to the rocket. The rocket starts from rest on the ground at time \(t = 0\). At time \(t\) seconds, \(t \leqslant 2\), the speed of the rocket is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the mass of the rocket including its fuel is \(m _ { 0 } ( 5 - 2 t ) \mathrm { kg }\). It is assumed that air resistance is negligible and the acceleration due to gravity is constant.

1. Show that, for \(t \leqslant 2\) $$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 2 u } { 5 - 2 t } - 9.8$$
2. Find the speed of the rocket at the instant when all of its fuel has been burnt.

6. A small object \(P\), of mass \(m _ { 0 }\), is projected vertically upwards from the ground with speed \(U\). As \(P\) moves upwards it picks up droplets of moisture from the atmosphere. The droplets are at rest immediately before they are picked up. In a model of the motion, \(P\) is modelled as a particle, air resistance is assumed to be negligible and the acceleration due to gravity is assumed to have the constant value of \(g\). When \(P\) is at a height \(x\) above the ground, the combined mass of \(P\) and the moisture is \(m _ { 0 } ( 1 + k x )\), where \(k\) is a constant, and the speed of \(P\) is \(v\).

1. Show that, while \(P\) is moving upwards $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( v ^ { 2 } \right) + \frac { 2 k v ^ { 2 } } { ( 1 + k x ) } = - 2 g$$ The general solution of this differential equation is given by \(v ^ { 2 } = \frac { A } { ( 1 + k x ) ^ { 2 } } - \frac { 2 g } { 3 k } ( 1 + k x )\),
   where \(A\) is an arbitrary constant. Given that \(U = \sqrt { 2 g h }\) and \(k = \frac { 7 } { 3 h }\)
2. find, in terms of \(h\), the height of \(P\) above the ground when \(P\) first comes to rest.

2 A particle P of mass \(m\) travels in a straight line on a smooth horizontal surface.
At time \(t , \mathrm { P }\) is a distance \(x\) from a fixed point O and is moving with speed \(v\) away from O . A horizontal force of magnitude \(3 m t\) acts on P , in a direction away from O .

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 3 t\).
2. Verify that the general solution of this differential equation is \(x = \frac { 1 } { 2 } t ^ { 3 } + A t + k\), where \(A\) and \(k\) are constants.
3. Given that \(x = 6\) and \(v = 12\) when \(t = 1\), find the values of \(A\) and \(k\).

14

1. Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } - 2 y = 12 e ^ { - x }\). You are given that \(y\) tends to zero as \(x\) tends to infinity, and that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 0\) when \(x = 0\).
2. Find the exact value of \(x\) for which \(y = 0\).

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.