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

7 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model.

2 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?

3 Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced. The population \(x\), in thousands, of red squirrels is modelled by the equation $$x = \frac { a } { 1 + k t }$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0 , x = 2.5\).

1. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { k x ^ { 2 } } { a }\).
2. Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate \(a\) and \(k\).
3. What is the long-term population of red squirrels predicted by this model? The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = 2 y - y ^ { 2 }$$ When \(t = 0 , y = 1\).
4. Express \(\frac { 1 } { 2 y - y ^ { 2 } }\) in partial fractions.
5. Hence show by integration that \(\ln \left( \frac { y } { 2 y } \right) = 2 t\). Show that \(y = \frac { 2 } { 1 + \mathrm { e } ^ { - 2 t } }\).
6. What is the long-term population of grey squirrels predicted by this model?

1 A drug is administered by an intravenous drip. The concentration, \(x\), of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after \(t\) hours is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right)$$ where \(0 \leqslant x < 1\), and \(k\) is a positive constant. Initially, \(x = 0\).

1. Express \(\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }\) in partial fractions.
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2. Hence solve the differential equation to show that \(\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }\).
3. After 1 hour the drug concentration reaches \(75 \%\) of its maximum value and so \(x = 0.75\). Find the value of \(k\), and the time taken for the drug concentration to reach \(90 \%\) of its maximum value.
4. Rearrange the equation in part (ii) to show that \(x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }\). Verify that in the long term the drug concentration approaches its maximum value.

3 Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \begin{figure}[h] \end{figure} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A \sqrt { h }$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.

1. Verify that the solution to this differential equation is $$h = \left( 1 - \frac { 1 } { 2 } A t \right) ^ { 2 } .$$ The water stops leaking when \(h = 0\). This occurs after 20 seconds.
2. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m . Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_235_455_1425_820} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure} For this barrel, $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - B \frac { \sqrt { h } } { ( 1 + h ) ^ { 2 } } ,$$ where \(B\) is a positive constant. When \(t = 0 , h = 1\).
3. Solve this differential equation, and hence show that $$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
4. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m .

4 The motion of a particle is modelled by the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).

1. Solve the differential equation to show that \(v ^ { 2 } = 20 - 4 x ^ { 2 }\). Now consider motion for which \(x = \cos 2 t + 2 \sin 2 t\), where \(x\) is the displacement from a fixed point at time \(t\).
2. Verify that, when \(t = 0 , x = 1\). Use the fact that \(v = \frac { \mathrm { d } x } { \mathrm {~d} t }\) to verify that when \(t = 0 , v = 4\).
3. Express \(x\) in the form \(R \cos ( 2 t - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v ^ { 2 } = 20 - 4 x ^ { 2 }\).
4. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value.

5 The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(\pounds V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).

1. Express the model as a differential equation. Verify by differentiation that \(V = \left( \frac { 1 } { 2 } k t + c \right) ^ { 2 }\), where \(c\) is an arbitrary constant, satisfies this differential equation.
2. The value of the company's sales in its first year is \(\pounds 10000\), and the total value of the sales in the first two years is \(\pounds 40000\). Find \(V\) in terms of \(t\).

1 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( x + 1 ) }\), given that when \(x = 1 , y = 1\). Your answer should express \(y\) explicitly in terms of \(x\).

2 Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x \mathrm {~cm}\), and the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 1 } { 3 } x ^ { 3 }\). The rate at which water is lost is proportional to \(x\), so that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x\), where \(k\) is a constant.

1. Show that \(x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k\). Initially, the depth of water in the container is 10 cm .
2. Show by integration that \(x = \sqrt { 100 - 2 k t }\).
3. Given that the container empties after 50 seconds, find \(k\). Once the container is empty, water is poured into it at a constant rate of \(1 \mathrm {~cm} ^ { 3 }\) per second. The container continues to lose water as before.
4. Show that, \(t\) seconds after starting to pour the water in, \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }\).
5. Show that \(\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }\). Hence solve the differential equation in part (iv) to show that $$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
6. Show that the depth cannot reach 1 cm .

3 A curve satisfies the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } y\), and passes through the point \(( 1,1 )\). Find \(y\) in terms of \(x\).

4 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).

1. Find \(v\) in terms of \(t\).
2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
5. According to this model, what is the speed of the skydiver in the long term?

5 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model.

6

1. The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating \(x\), the number of bacteria, to the time \(t\).
2. In another colony, the number of bacteria, \(y\), after time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 10000 } { \sqrt { y } }$$ Find \(y\) in terms of \(t\), given that \(y = 900\) when \(t = 0\). Hence find the number of bacteria after 10 minutes.

2 A curve has equation $$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 } ,$$ where \(k\) is a positive constant.

1. Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
2. Hence or otherwise show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }\).
3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-2_658_1070_861_567} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
4. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1 , k = 3\) and \(k = 4\). On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
5. Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
6. Solve this differential equation. Given that the path of the stream passes through the point \(( 2,1 )\), show that its equation is \(y = \frac { x ^ { 4 } } { 16 }\).

3.Given that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( u \sqrt { } x ) = \frac { \mathrm { d } u } { \mathrm {~d} x } \times \frac { \mathrm { d } ( \sqrt { } x ) } { \mathrm { d } x } , \quad 0 < x < \frac { 1 } { 2 }$$ where \(u\) is a function of \(x\) ,and that \(u = 4\) when \(x = \frac { 3 } { 8 }\) ,find \(u\) in terms of \(x\) .
(9)

4.The function \(\mathrm { h } ( x )\) has domain \(\mathbb { R }\) and range \(\mathrm { h } ( x ) > 0\) ,and satisfies $$\sqrt { \int \mathrm { h } ( x ) \mathrm { d } x } = \int \sqrt { \mathrm { h } ( x ) } \mathrm { d } x$$

1. By substituting \(\mathrm { h } ( x ) = \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 }\) ,show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( y + c ) ,$$ where \(c\) is constant.
2. Hence find a general expression for \(y\) in terms of \(x\) .
3. Given that \(\mathrm { h } ( 0 ) = 1\) ,find \(\mathrm { h } ( x )\) .

13. \begin{figure}[h] \end{figure} Figure 4 shows a hemispherical bowl containing some water.
At \(t\) seconds, the height of the water is \(h \mathrm {~cm}\) and the volume of the water is \(V \mathrm {~cm} ^ { 3 }\), where $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 30 - h ) , \quad 0 < h \leqslant 10$$ The water is leaking from a hole in the bottom of the bowl. Given that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - \frac { 1 } { 10 } V\)

1. show that \(\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { h ( 30 - h ) } { 30 ( 20 - h ) }\)
2. Write \(\frac { 30 ( 20 - h ) } { h ( 30 - h ) }\) in partial fraction form. Given that \(h = 10\) when \(t = 0\),
3. use your answers to parts (a) and (b) to find the time taken for the height of the water to fall to 5 cm . Give your answer in seconds to 2 decimal places.

10 In a chemical reaction, a compound \(X\) is formed from two compounds \(Y\) and \(Z\).
The masses in grams of \(X , Y\) and \(Z\) present at time \(t\) seconds after the start of the reaction are \(x , 10 - x\) and \(20 - x\) respectively. At any time the rate of formation of \(X\) is proportional to the product of the masses of \(Y\) and \(Z\) present at the time. When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.01 ( 10 - x ) ( 20 - x ) .$$
2. Solve this differential equation and obtain an expression for \(x\) in terms of \(t\).
3. State what happens to the value of \(x\) when \(t\) becomes large.
   

8 \includegraphics[max width=\textwidth, alt={}, center]{c940af95-e291-402a-856c-9090d13163d5-4_538_702_264_719} The diagram shows the curve with equation $$y = \frac { 6 } { \sqrt { x } } - 3$$ The point \(P\) has coordinates \(( 0 , p )\). The shaded region is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = p\). The shaded region is rotated completely about the \(y\)-axis to form a solid of volume \(V\).

1. Show that \(V = 16 \pi \left( 1 - \frac { 27 } { ( p + 3 ) ^ { 3 } } \right)\).
2. It is given that \(P\) is moving along the \(y\)-axis in such a way that, at time \(t\), the variables \(p\) and \(t\) are related by $$\frac { \mathrm { d } p } { \mathrm {~d} t } = \frac { 1 } { 3 } p + 1 .$$ Find the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\) at the instant when \(p = 9\).

9 A liquid is being heated in an oven maintained at a constant temperature of \(160 ^ { \circ } \mathrm { C }\). It may be assumed that the rate of increase of the temperature of the liquid at any particular time \(t\) minutes is proportional to \(160 - \theta\), where \(\theta ^ { \circ } \mathrm { C }\) is the temperature of the liquid at that time.

1. Write down a differential equation connecting \(\theta\) and \(t\). When the liquid was placed in the oven, its temperature was \(20 ^ { \circ } \mathrm { C }\) and 5 minutes later its temperature had risen to \(65 ^ { \circ } \mathrm { C }\).
2. Find the temperature of the liquid, correct to the nearest degree, after another 5 minutes. 4

10

1. Express \(\frac { 1 } { ( 3 - x ) ( 6 - x ) }\) in partial fractions.
2. In a chemical reaction, the amount \(x\) grams of a substance at time \(t\) seconds is related to the rate at which \(x\) is changing by the equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 3 - x ) ( 6 - x )$$ where \(k\) is a constant. When \(t = 0 , x = 0\) and when \(t = 1 , x = 1\).
   1. Show that \(k = \frac { 1 } { 3 } \ln \frac { 5 } { 4 }\).
   2. Find the value of \(x\) when \(t = 2\).

9 Paraffin is stored in a tank with a horizontal base. At time \(t\) minutes, the depth of paraffin in the tank is \(x \mathrm {~cm}\). When \(t = 0 , x = 72\). There is a tap in the side of the tank through which the paraffin can flow. When the tap is opened, the flow of the paraffin is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 4 ( x - 8 ) ^ { \frac { 1 } { 3 } }$$

1. How long does it take for the level of paraffin to fall from a depth of 72 cm to a depth of 35 cm ?
2. The tank is filled again to its original depth of 72 cm of paraffin and the tap is then opened. The paraffin flows out until it stops. How long does this take?

10

1. Write down the derivative of \(\sqrt { y ^ { 2 } + 1 }\) with respect to \(y\).
2. Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( x - 1 ) \sqrt { y ^ { 2 } + 1 } } { x y }\) and that \(y = \sqrt { \mathrm { e } ^ { 2 } - 2 \mathrm { e } }\) when \(x = \mathrm { e }\),
   find a relationship between \(x\) and \(y\).

9 The temperature of a freezer is \(- 20 ^ { \circ } \mathrm { C }\). A container of a liquid is placed in the freezer. The rate at which the temperature, \(\theta ^ { \circ } \mathrm { C }\), of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$ where time \(t\) is in minutes and \(k\) is a positive constant.

1. Express \(\theta\) in terms of \(t , k\) and an arbitrary constant. Initially the temperature of the liquid in the container is \(40 ^ { \circ } \mathrm { C }\) and, at this instant, the liquid is cooling at a rate of \(3 ^ { \circ } \mathrm { C }\) per minute. The liquid freezes at \(0 ^ { \circ } \mathrm { C }\).
2. Find the value of \(k\) and find also the time it takes (to the nearest minute) for the liquid to freeze. The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is \(90 ^ { \circ } \mathrm { C }\). After 19 minutes its temperature is \(0 ^ { \circ } \mathrm { C }\).
3. Without any further calculation, explain what you can deduce about the value of \(k\) in this case.