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

7. The volume of a spherical balloon of radius \(r \mathrm {~cm}\) is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\). The volume of the balloon increases with time \(t\) seconds according to the formula $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } } , \quad t \geqslant 0$$
2. Using the chain rule, or otherwise, find an expression in terms of \(r\) and \(t\) for \(\frac { \mathrm { d } r } { \mathrm {~d} t }\).
3. Given that \(V = 0\) when \(t = 0\), solve the differential equation \(\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } }\), to obtain \(V\) in terms of \(t\).
4. Hence, at time \(t = 5\),
   1. find the radius of the balloon, giving your answer to 3 significant figures,
   2. show that the rate of increase of the radius of the balloon is approximately \(2.90 \times 10 ^ { - 2 } \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).

4.

1. Express \(\frac { 2 x - 1 } { ( x - 1 ) ( 2 x - 3 ) }\) in partial fractions.
2. Given that \(x \geqslant 2\), find the general solution of the differential equation $$( 2 x - 3 ) ( x - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x - 1 ) y$$
3. Hence find the particular solution of this differential equation that satisfies \(y = 10\) at \(x = 2\), giving your answer in the form \(y = \mathrm { f } ( x )\).
   

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of \(1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is \(4000 \mathrm {~cm} ^ { 2 }\).

1. Show that at time \(t\) seconds, the height \(h \mathrm {~cm}\) of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$ When \(h = 25\), water is leaking out of the hole at \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$ Using the substitution \(h = ( 20 - x ) ^ { 2 }\), or otherwise,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.

5.

1. Find \(\int \frac { 9 x + 6 } { x } \mathrm {~d} x , x > 0\).
2. Given that \(y = 8\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 9 x + 6 ) y ^ { \frac { 1 } { 3 } } } { x }$$ giving your answer in the form \(y ^ { 2 } = \mathrm { g } ( x )\). \section*{LU}

3.

1. Express \(\frac { 5 } { ( x - 1 ) ( 3 x + 2 ) }\) in partial fractions.
2. Hence find \(\int \frac { 5 } { ( x - 1 ) ( 3 x + 2 ) } \mathrm { d } x\), where \(x > 1\).
3. Find the particular solution of the differential equation $$( x - 1 ) ( 3 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 5 y , \quad x > 1$$ for which \(y = 8\) at \(x = 2\). Give your answer in the form \(y = \mathrm { f } ( x )\).

8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at \(3 ^ { \circ } \mathrm { C }\) and \(t\) minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is \(\theta ^ { \circ } \mathrm { C }\). The rate of change of the temperature of the water in the bottle is modelled by the differential equation, $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \frac { ( 3 - \theta ) } { 125 }$$

1. By solving the differential equation, show that, $$\theta = A \mathrm { e } ^ { - 0.008 t } + 3$$ where \(A\) is a constant. Given that the temperature of the water in the bottle when it was put in the refrigerator was \(16 ^ { \circ } \mathrm { C }\),
2. find the time taken for the temperature of the water in the bottle to fall to \(10 ^ { \circ } \mathrm { C }\), giving your answer to the nearest minute.

1. Given that \(y = 2\) at \(x = \frac { \pi } { 8 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y ^ { 2 } } { 2 \sin ^ { 2 } 2 x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-17_81_102_2649_1779}

Liquid is pouring into a container at a constant rate of \(20 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out at a rate proportional to the volume of liquid already in the container.

1. Explain why, at time \(t\) seconds, the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container satisfies the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 20 - k V$$ where \(k\) is a positive constant. The container is initially empty.
2. By solving the differential equation, show that $$V = A + B \mathrm { e } ^ { - k t }$$ giving the values of \(A\) and \(B\) in terms of \(k\). Given also that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 10\) when \(t = 5\),
3. find the volume of liquid in the container at 10 s after the start.
   

7. At time \(t\) seconds the length of the side of a cube is \(x \mathrm {~cm}\), the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\), and the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\). The surface area of the cube is increasing at a constant rate of \(8 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\).
Show that

1. \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\), where \(k\) is a constant to be found,
2. \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 2 V ^ { \frac { 1 } { 3 } }\). Given that \(V = 8\) when \(t = 0\),
3. solve the differential equation in part (b), and find the value of \(t\) when \(V = 16 \sqrt { } 2\).

8. A population growth is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = k P ,$$ where \(P\) is the population, \(t\) is the time measured in days and \(k\) is a positive constant.
Given that the initial population is \(P _ { 0 }\),

1. solve the differential equation, giving \(P\) in terms of \(P _ { 0 } , k\) and \(t\). Given also that \(k = 2.5\),
2. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\). In an improved model the differential equation is given as $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \lambda P \cos \lambda t$$ where \(P\) is the population, \(t\) is the time measured in days and \(\lambda\) is a positive constant.
   Given, again, that the initial population is \(P _ { 0 }\) and that time is measured in days,
3. solve the second differential equation, giving \(P\) in terms of \(P _ { 0 } , \lambda\) and \(t\). Given also that \(\lambda = 2.5\),
4. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\) for the first time, using the improved model.

7.

1. Express \(\frac { 2 } { 4 - y ^ { 2 } }\) in partial fractions.
2. Hence obtain the solution of $$2 \cot x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 4 - y ^ { 2 } \right)$$ for which \(y = 0\) at \(x = \frac { \pi } { 3 }\), giving your answer in the form \(\sec ^ { 2 } x = \mathrm { g } ( y )\).
   

8. \begin{figure}[h] \end{figure} Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m . Water is flowing into the tank at a constant rate of \(0.48 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). At time \(t\) minutes, the depth of the water in the tank is \(h\) metres. There is a tap at a point \(T\) at the bottom of the tank. When the tap is open, water leaves the tank at a rate of \(0.6 \pi h \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\).

1. Show that \(t\) minutes after the tap has been opened $$75 \frac { \mathrm {~d} h } { \mathrm {~d} t } = ( 4 - 5 h )$$ When \(t = 0 , h = 0.2\)
2. Find the value of \(t\) when \(h = 0.5\)

1. Given that \(y = 2\) at \(x = \frac { \pi } { 4 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { y \cos ^ { 2 } x }$$

6. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta ^ { \circ } \mathrm { C }\). The rate of increase of the temperature of the water at any time \(t\) is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) , \quad \theta \leqslant 100$$ where \(\lambda\) is a positive constant. Given that \(\theta = 20\) when \(t = 0\),

1. solve this differential equation to show that $$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$ When the temperature of the water reaches \(100 ^ { \circ } \mathrm { C }\), the kettle switches off.
2. Given that \(\lambda = 0.01\), find the time, to the nearest second, when the kettle switches off.

7. The rate of increase of the number, \(N\), of fish in a lake is modelled by the differential equation $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { ( k t - 1 ) ( 5000 - N ) } { t } \quad t > 0 , \quad 0 < N < 5000$$ In the given equation, the time \(t\) is measured in years from the start of January 2000 and \(k\) is a positive constant.

1. By solving the differential equation, show that $$N = 5000 - A t \mathrm { e } ^ { - k t }$$ where \(A\) is a positive constant. After one year, at the start of January 2001, there are 1200 fish in the lake. After two years, at the start of January 2002, there are 1800 fish in the lake.
2. Find the exact value of the constant \(A\) and the exact value of the constant \(k\).
3. Hence find the number of fish in the lake after five years. Give your answer to the nearest hundred fish. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \section*{Question 7 continued} \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

4. The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 5 } { 2 } x , \quad t \geqslant 0$$ where \(x\) is the mass of the substance measured in grams and \(t\) is the time measured in days.
Given that \(x = 60\) when \(t = 0\),

1. solve the differential equation, giving \(x\) in terms of \(t\). You should show all steps in your working and give your answer in its simplest form.
2. Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

7. \begin{figure}[h] \end{figure} Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole \(P\) on the side of the tank. At time \(t\) minutes after the leaking starts, the height of water in the tank is \(h \mathrm {~cm}\). The height \(h \mathrm {~cm}\) of the water in the tank satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = k ( h - 9 ) ^ { \frac { 1 } { 2 } } , \quad 9 < h \leqslant 200$$ where \(k\) is a constant. Given that, when \(h = 130\), the height of the water is falling at a rate of 1.1 cm per minute,

1. find the value of \(k\). Given that the tank was full of water when the leaking started,
2. solve the differential equation with your value of \(k\), to find the value of \(t\) when \(h = 50\)

1. Given that \(y = 2\) when \(x = - \frac { \pi } { 8 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } } { 3 \cos ^ { 2 } 2 x } \quad - \frac { 1 } { 2 } < x < \frac { 1 } { 2 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

8. A circular stain grows in such a way that the rate of increase of its radius is inversely proportional to the square of the radius. Given that the area of the stain at time \(t\) seconds is \(A \mathrm {~cm} ^ { 2 }\),

1. show that \(\frac { \mathrm { d } A } { \mathrm {~d} t } \propto \frac { 1 } { \sqrt { A } }\).
   (6) Another stain, which is growing more quickly, has area \(S \mathrm {~cm} ^ { 2 }\) at time \(t\) seconds. It is given that $$\frac { \mathrm { d } S } { \mathrm {~d} t } = \frac { 2 \mathrm { e } ^ { 2 t } } { \sqrt { S } }$$ Given that, for this second stain, \(S = 9\) at time \(t = 0\),
2. solve the differential equation to find the time at which \(S = 16\). Give your answer to 2 significant figures. \section*{END}

8.

1. Given that \(y = 1\) at \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x y ^ { \frac { 1 } { 3 } } } { \mathrm { e } ^ { 2 x } } \quad y \geqslant 0$$ giving your answer in the form \(y ^ { 2 } = \mathrm { g } ( x )\).
2. Hence find the equation of the horizontal asymptote to the curve with equation \(y ^ { 2 } = \mathrm { g } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-27_2644_1840_118_111} \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-29_2646_1838_121_116}
   

1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

The temperature, \(\theta ^ { \circ } \mathrm { C }\), of a car engine, \(t\) minutes after the engine is turned off, is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 15 ) ^ { 2 }$$ where \(k\) is a constant.
Given that the temperature of the car engine

• is \(85 ^ { \circ } \mathrm { C }\) at the instant the engine is turned off
• is \(40 ^ { \circ } \mathrm { C }\) exactly 10 minutes after the engine is turned off
  1. solve the differential equation to show that, according to the model

$$\theta = \frac { a t + b } { c t + d }$$ where \(a , b , c\) and \(d\) are integers to be found.

Hence find, according to the model, the time taken for the temperature of the car engine to reach \(20 ^ { \circ } \mathrm { C }\). Give your answer to the nearest minute.

9. Bacteria are growing on the surface of a dish in a laboratory. The area of the dish, \(A \mathrm {~cm} ^ { 2 }\), covered by the bacteria, \(t\) days after the bacteria start to grow, is modelled by the differential equation $$\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { A ^ { \frac { 3 } { 2 } } } { 5 t ^ { 2 } } \quad t > 0$$ Given that \(A = 2.25\) when \(t = 3\)

1. show that $$A = \left( \frac { p t } { q t + r } \right) ^ { 2 }$$ where \(p , q\) and \(r\) are integers to be found. According to the model, there is a limit to the area that will be covered by the bacteria.
2. Find the value of this limit. \includegraphics[max width=\textwidth, alt={}, center]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-31_2255_50_314_34}
   

   VIIV SIHI NI JIIIM ION OC

   VIAV SIHI NI I II M I I O N OC

   VAYV SIHI NI JIIIM ION OO