AQA FP3 (Further Pure Mathematics 3) 2007 June

1

1. Find the value of the constant \(k\) for which \(k x ^ { 2 } \mathrm { e } ^ { 5 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 10 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 25 y = 6 \mathrm { e } ^ { 5 x }$$
2. Hence find the general solution of this differential equation.

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

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.
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 \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.

3 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \sec x$$ given that \(y = 3\) when \(x = 0\).

4

1. Show that \(( \cos \theta + \sin \theta ) ^ { 2 } = 1 + \sin 2 \theta\).
2. A curve has cartesian equation $$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = ( x + y ) ^ { 4 }$$ Given that \(r \geqslant 0\), show that the polar equation of the curve is $$r = 1 + \sin 2 \theta$$
3. The curve with polar equation $$r = 1 + \sin 2 \theta , \quad - \pi \leqslant \theta \leqslant \pi$$ is shown in the diagram.
   \includegraphics[max width=\textwidth, alt={}, center]{f90167c3-2ffd-464a-b2d2-9f86a8d64887-3_389_611_1062_708}
   1. Find the two values of \(\theta\) for which \(r = 0\).
   2. Find the area of one of the loops.

5

1. A differential equation is given by $$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$ Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + x$$ transforms this differential equation into $$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$ (4 marks)
2. Find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

6

1. The function f is defined by $$\mathrm { f } ( x ) = \ln \left( 1 + \mathrm { e } ^ { x } \right)$$ Use Maclaurin's theorem to show that when \(\mathrm { f } ( x )\) is expanded in ascending powers of \(x\) :
   1. the first three terms are $$\ln 2 + \frac { 1 } { 2 } x + \frac { 1 } { 8 } x ^ { 2 }$$
   2. the coefficient of \(x ^ { 3 }\) is zero.
2. Hence write down the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right)\).
3. Use the series expansion $$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \ldots$$ to write down the first three terms in the expansion, in ascending powers of \(x\), of \(\ln \left( 1 - \frac { x } { 2 } \right)\).
4. Use your answers to parts (b) and (c) to find $$\lim _ { x \rightarrow 0 } \left[ \frac { \ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right) + \ln \left( 1 - \frac { x } { 2 } \right) } { x - \sin x } \right]$$

7

1. Write down the value of $$\lim _ { x \rightarrow \infty } x \mathrm { e } ^ { - x }$$
2. Use the substitution \(u = x \mathrm { e } ^ { - x } + 1\) to find \(\int \frac { \mathrm { e } ^ { - x } ( 1 - x ) } { x \mathrm { e } ^ { - x } + 1 } \mathrm {~d} x\).
3. Hence evaluate \(\int _ { 1 } ^ { \infty } \frac { 1 - x } { x + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing the limiting process used.