Separable with partial fractions

8

1. Using partial fractions, find $$\int \frac { 1 } { y ( 4 - y ) } \mathrm { d } y$$
2. Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 4 - y ) ,$$ obtaining an expression for \(y\) in terms of \(x\).
3. State what happens to the value of \(y\) if \(x\) becomes very large and positive.

9 The variables \(x\) and \(t\) satisfy the differential equation \(5 \frac { \mathrm {~d} x } { \mathrm {~d} t } = ( 20 - x ) ( 40 - x )\). It is given that \(x = 10\) when \(t = 0\).

1. Using partial fractions, solve the differential equation, obtaining an expression for \(x\) in terms of \(t\). [9]
2. State what happens to the value of \(x\) when \(t\) becomes large.

10 The variables \(x\) and \(t\) satisfy the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 + 2 x )\), and \(x = 1\) when \(t = 0\).
Using partial fractions, solve the differential equation, obtaining an expression for \(t\) in terms of \(x\).
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

1. The number of goats on an island is being monitored.

When monitoring began there were 3000 goats on the island.
In a simple model, the number of goats, \(x\), in thousands, is modelled by the equation $$x = \frac { k ( 9 t + 5 ) } { 4 t + 3 }$$ where \(k\) is a constant and \(t\) is the number of years after monitoring began.

1. Show that \(k = 1.8\)
2. Hence calculate the long-term population of goats predicted by this model. In a second model, the number of goats, \(x\), in thousands, is modelled by the differential equation $$3 \frac { \mathrm {~d} x } { \mathrm {~d} t } = x ( 9 - 2 x )$$
3. Write \(\frac { 3 } { x ( 9 - 2 x ) }\) in partial fraction form.
4. Solve the differential equation with the initial condition to show that $$x = \frac { 9 } { 2 + \mathrm { e } ^ { - 3 t } }$$
5. Find the long-term population of goats predicted by this second model.

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.

4 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?

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.

7 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?

14. a. Express \(\frac { 1 } { ( 3 - x ) ( 1 - x ) }\) in partial fractions.
(2) A scientist is studying the mass of a substance in a laboratory.
The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation $$2 \frac { d x } { d t } = ( 3 - x ) ( 1 - x ) \quad t \geq 0,0 \leq x < 1$$ Given that when \(t = 0 , x = 0\)
b. solve the differential equation and show that the solution can be written as $$x = \frac { 3 \left( e ^ { t } - 1 \right) } { 3 e ^ { t } - 1 }$$ c. Find the mass, \(x\) grams, which has formed 2 seconds after the start of the reaction. Give your answer correct to 3 significant figures.
d. Find the limiting value of \(x\) as \(t\) increases.

10

1. Express \(\frac { 1 } { ( 4 x + 1 ) ( x + 1 ) }\) in partial fractions.
2. A curve passes through the point \(( 0,2 )\) and satisfies the differential equation \(\frac { d y } { d x } = \frac { y } { ( 4 x + 1 ) ( x + 1 ) }\),
   for \(x > - \frac { 1 } { 4 }\).
   Show by integration that \(\mathrm { y } = \mathrm { A } \left( \frac { 4 \mathrm { x } + 1 } { \mathrm { x } + 1 } \right) ^ { \mathrm { B } }\) where \(A\) and \(B\) are constants to be determined.