AQA FP3 (Further Pure Mathematics 3) 2009 January

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { x ^ { 2 } + y ^ { 2 } } { x + y }$$ and $$y ( 1 ) = 3$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\).
2. Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\), giving your answer to four decimal places.

2

1. Show that \(\frac { 1 } { x ^ { 2 } }\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 2 } { x } y = x$$
2. Hence find the general solution of this differential equation, giving your answer in the form \(y = \mathrm { f } ( x )\).

3 The diagram shows a sketch of a loop, the pole \(O\) and the initial line.
\includegraphics[max width=\textwidth, alt={}, center]{f4fdffc7-5647-4462-a983-1564d4e76a4d-3_305_553_383_740} The polar equation of the loop is $$r = ( 2 + \cos \theta ) \sqrt { \sin \theta } , \quad 0 \leqslant \theta \leqslant \pi$$ Find the area enclosed by the loop.

4

1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\), where \(c\) is an arbitrary constant.
2. Hence evaluate \(\int _ { 0 } ^ { 1 } \ln x \mathrm {~d} x\), showing the limiting process used.

5 The diagram shows a sketch of a curve \(C\), the pole \(O\) and the initial line.
\includegraphics[max width=\textwidth, alt={}, center]{f4fdffc7-5647-4462-a983-1564d4e76a4d-3_301_668_1644_689} The curve \(C\) has polar equation $$r = \frac { 2 } { 3 + 2 \cos \theta } , \quad 0 \leqslant \theta \leqslant 2 \pi$$

1. Verify that the point \(L\) with polar coordinates ( \(2 , \pi\) ) lies on \(C\).
2. The circle with polar equation \(r = 1\) intersects \(C\) at the points \(M\) and \(N\).
   1. Find the polar coordinates of \(M\) and \(N\).
   2. Find the area of triangle \(L M N\).
3. Find a cartesian equation of \(C\), giving your answer in the form \(9 y ^ { 2 } = \mathrm { f } ( x )\).

6 The function f is defined by \(\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } ( 1 + 3 x ) ^ { - \frac { 2 } { 3 } }\).

1. 1. Use the series expansion for \(\mathrm { e } ^ { x }\) to write down the first four terms in the series expansion of \(\mathrm { e } ^ { 2 x }\).
   2. Use the binomial series expansion of \(( 1 + 3 x ) ^ { - \frac { 2 } { 3 } }\) and your answer to part (a)(i) to show that the first three non-zero terms in the series expansion of \(\mathrm { f } ( x )\) are \(1 + 3 x ^ { 2 } - 6 x ^ { 3 }\).
2. 1. Given that \(y = \ln ( 1 + 2 \sin x )\), find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   2. By using Maclaurin's theorem, show that, for small values of \(x\), $$\ln ( 1 + 2 \sin x ) \approx 2 x - 2 x ^ { 2 }$$
3. Find $$\lim _ { x \rightarrow 0 } \frac { 1 - \mathrm { f } ( x ) } { x \ln ( 1 + 2 \sin x ) }$$

7

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
3. Find the general solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10\).
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 8\) when \(x = 1\).