Separable variables - standard (polynomial/exponential x-side)

8

1. A water tank has a height of 2 metres. The depth of the water in the tank is \(h\) metres at time \(t\) minutes after water begins to enter the tank. The rate at which the depth of the water in the tank increases is proportional to the difference between the height of the tank and the depth of the water. Write down a differential equation in the variables \(h\) and \(t\) and a positive constant \(k\).
   (You are not required to solve your differential equation.)
2. 1. Another water tank is filling in such a way that \(t\) minutes after the water is turned on, the depth of the water, \(x\) metres, increases according to the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 15 x \sqrt { 2 x - 1 } }$$ The depth of the water is 1 metre when the water is first turned on.
      Solve this differential equation to find \(t\) as a function of \(x\).
   2. Calculate the time taken for the depth of the water in the tank to reach 2 metres, giving your answer to the nearest 0.1 of a minute.
      (l mark)

8

1. A pond is initially empty and is then filled gradually with water. After \(t\) minutes, the depth of the water, \(x\) metres, satisfies the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \sqrt { 4 + 5 x } } { 5 ( 1 + t ) ^ { 2 } }$$ Solve this differential equation to find \(x\) in terms of \(t\).
2. Another pond is gradually filling with water. After \(t\) minutes, the surface of the water forms a circle of radius \(r\) metres. The rate of change of the radius is inversely proportional to the area of the surface of the water.
   1. Write down a differential equation, in the variables \(r\) and \(t\) and a constant of proportionality, which represents how the radius of the surface of the water is changing with time.
      (You are not required to solve your differential equation.)
   2. When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second.
      [0pt] [2 marks]
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1. Given that \(y = - 2\) when \(x = 1\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } \sqrt { x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

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

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

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 \(y = \frac { 1 } { 2 } \left( 15 - \frac { 8 } { x } + \frac { 1 } { x ^ { 2 } } \right)\).

8

1. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 150 \cos 2 t } { x }$$ given that \(x = 20\) when \(t = \frac { \pi } { 4 }\), giving your solution in the form \(x ^ { 2 } = \mathrm { f } ( t )\). (6 marks)
2. The oscillations of a 'baby bouncy cradle' are modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 150 \cos 2 t } { x }$$ where \(x \mathrm {~cm}\) is the height of the cradle above its base \(t\) seconds after the cradle begins to oscillate. Given that the cradle is 20 cm above its base at time \(t = \frac { \pi } { 4 }\) seconds, find:
   1. the height of the cradle above its base 13 seconds after it starts oscillating, giving your answer to the nearest millimetre;
   2. the time at which the cradle will first be 11 cm above its base, giving your answer to the nearest tenth of a second.
      (2 marks)

11 Solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \sec y\) given that \(y = \frac { \pi } { 6 }\) when \(x = 4\) giving your answer in the form \(y = \mathrm { f } ( x )\).

7 Solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \sec y\) given that \(y = \frac { \pi } { 6 }\) when \(x = 4\) giving your answer in the form \(y = \mathrm { f } ( x )\).

6 Find the solution of the differential equation $$x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x + 1$$ given that \(y = 3\) when \(x = 1\). Give your answer in the form \(y = \mathrm { f } ( x )\).

7 Solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \sec y\) given that \(y = \frac { \neq } { 6 }\) when \(x = 4\) giving your answer in the form \(y = \mathrm { f } ( x )\).

The variables \(x\) and \(y\) satisfy the differential equation $$(1 - \cos x)\frac{dy}{dx} = y \sin x.$$ It is given that \(y = 4\) when \(x = \pi\).

1. Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\). [6]
2. Sketch the graph of \(y\) against \(x\) for \(0 < x < 2\pi\). [1]

Given that \(x = 1\) when \(t = 0\), solve the differential equation $$\frac{dx}{dt} = \frac{1}{x} - \frac{x}{4},$$ obtaining an expression for \(x^2\) in terms of \(t\). [7]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{dy}{dx} = 4 \cos^2 y \tan x,$$ for \(0 \leqslant x < \frac{1}{2}\pi\), and \(x = 0\) when \(y = \frac{1}{4}\pi\). Solve this differential equation and find the value of \(x\) when \(y = \frac{1}{8}\pi\). [8]

It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).

1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]

A forest is burning so that, \(t\) hours after the start of the fire, the area burnt is \(A\) hectares. It is given that, at any instant, the rate at which this area is increasing is proportional to \(A^2\).

1. Write down a differential equation which models this situation. [2]
2. After 1 hour, 1000 hectares have been burnt; after 2 hours, 2000 hectares have been burnt. Find after how many hours 3000 hectares have been burnt. [6]

A curve satisfies the differential equation \(\frac{dy}{dx} = 3x^2y\), and passes through the point \((1, 1)\). Find \(y\) in terms of \(x\). [4]

A mathematician is selling goods at a car boot sale. She believes that the rate at which she makes sales depends on the length of time since the start of the sale, \(t\) hours, and the total value of sales she has made up to that time, £\(x\). She uses the model $$\frac{dx}{dt} = \frac{k(5-t)}{x},$$ where \(k\) is a constant. Given that after two hours she has made sales of £96 in total,

1. solve the differential equation and show that she made £72 in the first hour of the sale. [8]

The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than £10 per hour.

1. Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left. [4]

The height \(x\) metres, of a column of water in a fountain display satisfies the differential equation \(\frac{dx}{dt} = \frac{8\sin 2t}{3\sqrt{x}}\), where \(t\) is the time in seconds after the display begins.

1. Solve the differential equation, given that initially the column of water has zero height. Express your answer in the form \(x = f(t)\) [7 marks]
2. Find the maximum height of the column of water, giving your answer to the nearest cm. [1 mark]

Solve the differential equation $$\frac{dt}{dx} = \frac{\ln x}{x^2 t} \quad \text{for } x > 0$$ given \(x = 1\) when \(t = 2\) Write your answer in the form \(t^2 = f(x)\) [7 marks]