OCR MEI C4 (Core Mathematics 4)

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.