1.08k Separable differential equations: dy/dx = f(x)g(y)

3 A model car of mass 2 kg moves from rest along a horizontal straight path. After time \(t \mathrm {~s}\), the velocity of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The power, \(P \mathrm {~W}\), developed by the engine is initially modelled by \(P = 2 v ^ { 3 } + 4 v\). The car is subject to a resistance force of magnitude \(6 v \mathrm {~N}\).

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = ( 1 - v ) ( 2 - v )\) and hence show that \(t = \ln \frac { 2 - v } { 2 ( 1 - v ) }\).
2. Hence express \(v\) in terms of \(t\). Once the power reaches 4.224 W it remains at this constant value with the resistance force still acting.
3. Verify that the power of 4.224 W is reached when \(v = 0.8\) and calculate the value of \(t\) at this instant.
4. Find \(v\) in terms of \(t\) for the motion at constant power. Deduce the limiting value of \(v\) as \(t \rightarrow \infty\).

2 On a building site, a pulley system is used for moving small amounts of material up to roof level. A light pulley, which can rotate freely, is attached with its axis horizontal to the top of some scaffolding. A light inextensible rope hangs over the pulley with a counterweight of mass \(m _ { 1 } \mathrm {~kg}\) attached to one end. Attached to the other end of the rope is a bag of negligible mass into which \(m _ { 2 } \mathrm {~kg}\) of roof tiles are placed, where \(m _ { 2 } < m _ { 1 }\). This situation is shown in Fig. 2. \begin{figure}[h] \end{figure} Initially the system is held at rest with the rope taut, the counterweight at the top of the scaffolding and the bag of tiles on the ground. When the counterweight is released, the bag ascends towards the top of the scaffolding. At time \(t \mathrm {~s}\) the velocity of the counterweight is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards. The counterweight is made from a bag of negligible mass filled with sand. At the moment the counterweight is released, this bag is accidentally ripped and after this time the sand drops out at a constant rate of \(\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }\).

1. Find the equation of motion for the counterweight while it still contains sand, and hence show that $$v = g t + \frac { 2 g m _ { 2 } } { \lambda } \ln \left( 1 - \frac { \lambda t } { m _ { 1 } + m _ { 2 } } \right) .$$
2. Given that the sand would run out after 10 seconds and that \(m _ { 2 } = \frac { 4 } { 5 } m _ { 1 }\), find the maximum velocity attained by the counterweight towards the ground. You may assume that the scaffolding is sufficiently high that the counterweight does not hit the ground before this velocity is reached.

4 A raindrop falls from rest through a stationary cloud. The raindrop has mass \(m\) and speed \(v\) when it has fallen a distance \(x\). You may assume that resistances to motion are negligible.

1. Derive the equation of motion $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } + v ^ { 2 } \frac { \mathrm {~d} m } { \mathrm {~d} x } = m g .$$ Initially the mass of the raindrop is \(m _ { 0 }\). Two different models for the mass of the raindrop are suggested.
   In the first model \(m = m _ { 0 } \mathrm { e } ^ { k _ { 1 } x }\), where \(k _ { 1 }\) is a positive constant.
2. Show that $$v ^ { 2 } = \frac { g } { k _ { 1 } } \left( 1 - \mathrm { e } ^ { - 2 k _ { 1 } x } \right) ,$$ and hence state, in terms of \(g\) and \(k _ { 1 }\), the terminal velocity of the raindrop according to this first model. In the second model \(m = m _ { 0 } \left( 1 + k _ { 2 } x \right)\), where \(k _ { 2 }\) is a positive constant.
3. By considering the expression obtained from differentiating \(v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 }\) with respect to \(x\), show that, for the second model, the equation of motion in part (i) may be written as $$\frac { d } { d x } \left[ v ^ { 2 } \left( 1 + k _ { 2 } x \right) ^ { 2 } \right] = 2 g \left( 1 + k _ { 2 } x \right) ^ { 2 } .$$ Hence find an expression for \(v ^ { 2 }\) in terms of \(g , k _ { 2 }\) and \(x\).
4. Suppose that the models give the same value for the speed of the raindrop at the instant when it has doubled its initial mass. Find the exact value of \(\frac { k _ { 1 } } { k _ { 2 } }\), giving your answer in the form \(\frac { a } { b }\) where \(a\) and \(b\) are integers. are integers. \section*{END OF QUESTION PAPER}

7 A box B of mass \(m \mathrm {~kg}\) is raised vertically by an engine working at a constant rate of \(k m g \mathrm {~W}\). Initially B is at rest. The speed of B when it has been raised a distance \(x \mathrm {~m}\) is denoted by \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v ^ { 2 } \frac { d v } { d x } = ( k - v ) g\).
2. Verify that \(\mathrm { gx } = \mathrm { k } ^ { 2 } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { v } } \right) - \mathrm { kv } - \frac { 1 } { 2 } \mathrm { v } ^ { 2 }\).
3. By using the work-energy principle, show that the time taken for B to reach a speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from rest is given by \(\frac { \mathrm { k } } { \mathrm { g } } \ln \left( \frac { \mathrm { k } } { \mathrm { k } - \mathrm { V } } \right) - \frac { \mathrm { V } } { \mathrm { g } }\).

A large aeroplane, of mass 360 tonnes, starts from rest at the beginning of a straight horizontal runway. The aeroplane produces a constant thrust of 980 kN and experiences a variable resistance to motion of magnitude \(\left( 80 + 0 \cdot 1 v ^ { 2 } \right) \mathrm { kN }\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the aeroplane after it has travelled \(x\) metres.

1. 1. Find the maximum speed that the aeroplane can attain.
   2. Show that \(v\) satisfies the differential equation $$3600 v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 9000 - v ^ { 2 } .$$
   (b) Find an expression for \(v ^ { 2 }\) in terms of \(x\).
2. Given that the aeroplane must achieve a speed of at least \(85 \mathrm {~ms} ^ { - 1 }\) to take off, determine the minimum length of the runway.
3. Explain why, according to this model, the aeroplane will not reach the speed found in (a)(i).

A ball of mass 0.4 kg is thrown vertically upwards from a point \(O\) with initial speed \(17 \mathrm {~ms} ^ { - 1 }\). When the ball is at a height of \(x \mathrm {~m}\) above \(O\) and its speed is \(v \mathrm {~ms} ^ { - 1 }\), the air resistance acting on the ball has magnitude \(0.01 v ^ { 2 } \mathrm {~N}\).

1. Show that, as the ball is ascending, \(v\) satisfies the differential equation $$40 v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \left( 392 + v ^ { 2 } \right)$$
2. Find an expression for \(v\) in terms of \(x\).
3. Calculate, correct to two decimal places, the greatest height of the ball.
4. State, with a reason, whether the speed of the ball when it returns to \(O\) is greater than \(17 \mathrm {~ms} ^ { - 1 }\), less than \(17 \mathrm {~ms} ^ { - 1 }\) or equal to \(17 \mathrm {~ms} ^ { - 1 }\).
   

3. A body, of mass 9 kg , is projected along a straight horizontal track with an initial speed of \(20 \mathrm {~ms} ^ { - 1 }\). At time \(t \mathrm {~s}\) the body experiences a resistance of magnitude \(( 0.2 + 0.03 v ) \mathrm { N }\) where \(v \mathrm {~ms} ^ { - 1 }\) is its speed.

1. Show that \(v\) satisfies the differential equation $$900 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( 20 + 3 v )$$
2. Find an expression for \(t\) in terms of \(v\).
3. Calculate, to the nearest second, the time taken for the body to come to rest.

1. A particle \(P\) is moving along the \(x\)-axis. At time \(t\) seconds, \(P\) has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\) direction and acceleration \(a \mathrm {~ms} ^ { - 2 }\) in the positive \(x\) direction.

In a model of the motion of \(P\) $$a = 4 - 3 v$$ When \(t = 0 , v = 0\)

1. Use integration to show that \(v = k \left( 1 - \mathrm { e } ^ { - 3 t } \right)\), where \(k\) is a constant to be found. When \(t = 0 , P\) is at the origin \(O\)
2. Find, in terms of \(t\) only, the distance of \(P\) from \(O\) at time \(t\) seconds.

16 In this question you must show detailed reasoning. Find the particular solution of the differential equation $$\frac { d y } { d x } = \frac { 9 y } { ( x - 1 ) ( x + 2 ) }$$ given that \(x = 2\) when \(y = 16\). \section*{END OF QUESTION PAPER}

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.

7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at \(25 \mathrm {~cm} ^ { 3 }\) per second. Water also leaks out of the tank at \(4 h ^ { 2 } \mathrm {~cm} ^ { 3 }\) per second, where \(h \mathrm {~cm}\) is the depth of the water after \(t\) seconds. Find the time taken for the water to reach a depth of 2 cm .

12 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } \sin 2 x } { 2 \cos ^ { 2 } 4 y - 1 }\).

1. Find an equation for the curve in the form \(\mathrm { f } ( y ) = g ( x )\). The curve passes through the point \(\left( \frac { 1 } { 4 } \pi , \frac { 1 } { 12 } \pi \right)\).
2. Find the smallest positive value of \(y\) for which \(x = 0\). \section*{END OF QUESTION PAPER}

7 As a spherical snowball melts its volume decreases. The rate of decrease of the volume of the snowball at any given time is modelled as being proportional to its volume at that time. Initially the volume of the snowball is \(500 \mathrm {~cm} ^ { 3 }\) and the rate of decrease of its volume is \(20 \mathrm {~cm} ^ { 3 }\) per hour.

1. Find the time that this model would predict for the snowball's volume to decrease to \(250 \mathrm {~cm} ^ { 3 }\).
2. Write down one assumption made when using this model.
3. Comment on how realistic this model would be in the long term.

8

1. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2 ( x - 6 ) ^ { \frac { 1 } { 2 } }$$ to find \(t\) in terms of \(x\), given that \(x = 70\) when \(t = 0\).
2. Liquid fuel is stored in a tank. At time \(t\) minutes, the depth of fuel in the tank is \(x \mathrm {~cm}\). Initially there is a depth of 70 cm of fuel in the tank. There is a tap 6 cm above the bottom of the tank. The flow of fuel out of the tank is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2 ( x - 6 ) ^ { \frac { 1 } { 2 } }$$
   1. Explain what happens when \(x = 6\).
   2. Find how long it will take for the depth of fuel to fall from 70 cm to 22 cm .

8

1. 1. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} t } = y \sin t\) to obtain \(y\) in terms of \(t\).
   2. Given that \(y = 50\) when \(t = \pi\), show that \(y = 50 \mathrm { e } ^ { - ( 1 + \cos t ) }\).
2. A wave machine at a leisure pool produces waves. The height of the water, \(y \mathrm {~cm}\), above a fixed point at time \(t\) seconds is given by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = y \sin t$$
   1. Given that this height is 50 cm after \(\pi\) seconds, find, to the nearest centimetre, the height of the water after 6 seconds.
   2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\) and hence verify that the water reaches a maximum height after \(\pi\) seconds.

8 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 \cos 3 x } { y }$$ given that \(y = 2\) when \(x = \frac { \pi } { 2 }\). Give your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

7

1. A differential equation is given by \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - k t \mathrm { e } ^ { \frac { 1 } { 2 } x }\), where \(k\) is a positive constant.
   1. Solve the differential equation.
   2. Hence, given that \(x = 6\) when \(t = 0\), show that \(x = - 2 \ln \left( \frac { k t ^ { 2 } } { 4 } + \mathrm { e } ^ { - 3 } \right)\).
      (3 marks)
2. The population of a colony of insects is decreasing according to the model \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - k t \mathrm { e } ^ { \frac { 1 } { 2 } x }\), where \(x\) thousands is the number of insects in the colony after time \(t\) minutes. Initially, there were 6000 insects in the colony. Given that \(k = 0.004\), find:
   1. the population of the colony after 10 minutes, giving your answer to the nearest hundred;
   2. the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.

7 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { y } \cos \left( \frac { x } { 3 } \right)\), given that \(y = 1\) when \(x = \frac { \pi } { 2 }\).
Write your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

9 A botanist is investigating the rate of growth of a certain species of toadstool. She observes that a particular toadstool of this type has a height of 57 millimetres at a time 12 hours after it begins to grow. She proposes the model \(h = A \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 4 } t } \right)\), where \(A\) is a constant, for the height \(h\) millimetres of the toadstool, \(t\) hours after it begins to grow.

1. Use this model to:
   1. find the height of the toadstool when \(t = 0\);
   2. show that \(A = 60\), correct to two significant figures.
2. Use the model \(h = 60 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 4 } t } \right)\) to:
   1. show that the time \(T\) hours for the toadstool to grow to a height of 48 millimetres is given by $$T = a \ln b$$ where \(a\) and \(b\) are integers;
   2. show that \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 15 - \frac { h } { 4 }\);
   3. find the height of the toadstool when it is growing at a rate of 13 millimetres per hour.
      (1 mark)

8

1. A cup of coffee is cooling down in a room. At time \(t\) minutes after the coffee is made, its temperature is \(x ^ { \circ } \mathrm { C }\), where $$x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }$$
   1. Find the temperature of the coffee when it is made.
   2. Find the temperature of the coffee 30 minutes after it is made.
   3. Find how long it will take for the coffee to cool down to \(60 ^ { \circ } \mathrm { C }\).
2. 1. Use integration to solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 1 } { 40 } ( x - 15 ) , \quad x > 15$$ given that \(x = 85\) when \(t = 0\), expressing \(t\) in terms of \(x\).
   2. Hence show that \(x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }\).

7 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x y ^ { 2 }$$ given that \(y = 1\) when \(x = 2\). Give your answer in the form \(y = \mathrm { f } ( x )\).

8 A disease is spreading through a colony of rabbits. There are 5000 rabbits in the colony. At time \(t\) hours, \(x\) is the number of rabbits infected. The rate of increase of the number of rabbits infected is proportional to the product of the number of rabbits infected and the number not yet infected.

1. 1. Formulate a differential equation for \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) in terms of the variables \(x\) and \(t\) and a constant of proportionality \(k\).
   2. Initially, 1000 rabbits are infected and the disease is spreading at a rate of 200 rabbits per hour. Find the value of the constant \(k\).
      (You are not required to solve your differential equation.)
2. The solution of the differential equation in this model is $$t = 4 \ln \left( \frac { 4 x } { 5000 - x } \right)$$
   1. Find the time after which 2500 rabbits will be infected, giving your answer in hours to one decimal place.
   2. Find, according to this model, the number of rabbits infected after 30 hours.

8

1. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 + 2 y } } { x ^ { 2 } }$$ given that \(y = 4\) when \(x = 1\).
2. Show that the solution can be written as \(y = \frac { 1 } { 2 } \left( 15 - \frac { 8 } { x } + \frac { 1 } { x ^ { 2 } } \right)\).

6

1. Express \(\frac { 2 } { x ^ { 2 } - 1 }\) in the form \(\frac { A } { x - 1 } + \frac { B } { x + 1 }\).
2. Hence find \(\int \frac { 2 } { x ^ { 2 } - 1 } \mathrm {~d} x\).
3. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y } { 3 \left( x ^ { 2 } - 1 \right) }\), given that \(y = 1\) when \(x = 3\). Show that the solution can be written as \(y ^ { 3 } = \frac { 2 ( x - 1 ) } { x + 1 }\).