Partial fractions for differential equations

7. (a) Express \(\frac { 2 } { 4 - y ^ { 2 } }\) in partial fractions.
(b) 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 \(\sec ^ { 2 } x = \mathrm { g } ( y )\).

6

1. Express \(\frac { 1 } { ( 2 x + 1 ) ( x + 1 ) }\) in partial fractions.
2. A curve passes through the point \(( 0,2 )\) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { ( 2 x + 1 ) ( x + 1 ) }$$ Show by integration that \(y = \frac { 4 x + 2 } { x + 1 }\). Section B (36 marks)

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.

10

1. Express \(\frac { 16 + 5 x - 2 x ^ { 2 } } { ( x + 1 ) ^ { 2 } ( x + 4 ) }\) in partial fractions.
2. It is given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( 16 + 5 x - 2 x ^ { 2 } \right) y } { ( x + 1 ) ^ { 2 } ( x + 4 ) }$$ and that \(y = \frac { 1 } { 256 }\) when \(x = 0\). Find the exact value of \(y\) when \(x = 2\). Give your answer in the form \(A \mathrm { e } ^ { n }\).

1. (a) Express \(\frac { 3 } { ( 2 x - 1 ) ( x + 1 ) }\) in partial fractions.

When chemical \(A\) and chemical \(B\) are mixed, oxygen is produced.
A scientist mixed these two chemicals and measured the total volume of oxygen produced over a period of time. The total volume of oxygen produced, \(V \mathrm {~m} ^ { 3 } , t\) hours after the chemicals were mixed, is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 3 V } { ( 2 t - 1 ) ( t + 1 ) } \quad V \geqslant 0 \quad t \geqslant k$$ where \(k\) is a constant.
Given that exactly 2 hours after the chemicals were mixed, a total volume of \(3 \mathrm {~m} ^ { 3 }\) of oxygen had been produced,
(b) solve the differential equation to show that $$V = \frac { 3 ( 2 t - 1 ) } { ( t + 1 ) }$$ The scientist noticed that

• there was a time delay between the chemicals being mixed and oxygen being produced
• there was a limit to the total volume of oxygen produced

Deduce from the model
(c) (i) the time delay giving your answer in minutes,
(ii) the limit giving your answer in \(\mathrm { m } ^ { 3 }\)

17

1. Express \(\frac { \left( x ^ { 2 } - 8 x + 9 \right) } { ( x + 1 ) ( x - 2 ) ^ { 2 } }\) in partial fractions.
2. Express \(y\) in terms of \(x\) given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y \left( x ^ { 2 } - 8 x + 9 \right) } { ( x + 1 ) ( x - 2 ) ^ { 2 } } \text { and } y = 16 \text { when } x = 3 .$$ \section*{END OF QUESTION PAPER}

8

1. Express \(\frac { 1 } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }\) in the form \(\frac { A } { 3 - 2 x } + \frac { B } { 1 - x } + \frac { C } { ( 1 - x ) ^ { 2 } }\).
   (4 marks)
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \sqrt { y } } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\), expressing your answer in the form $$y ^ { p } = q \ln [ \mathrm { f } ( x ) ] + \frac { x } { 1 - x }$$ where \(p\) and \(q\) are constants.

8

1. Express \(\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }\) in the form \(\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }\).
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\).
   Give your answer in the form \(\mathrm { f } ( y ) = \mathrm { g } ( x )\).
   [0pt] [7 marks]

6

1. Express \(\frac { 2 } { x ^ { 2 } - 1 }\) in the form \(\frac { A } { x - 1 } + \frac { B } { x + 1 }\).
2. Hence find \(\int \frac { 2 } { x ^ { 2 } - 1 } \mathrm {~d} x\).
3. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y } { 3 \left( x ^ { 2 } - 1 \right) }\), given that \(y = 1\) when \(x = 3\). Show that the solution can be written as \(y ^ { 3 } = \frac { 2 ( x - 1 ) } { x + 1 }\).