7
\end{array} \right) , \quad \mathbf { y } _ { 2 } = \left( \begin{array} { r }
2
- 3
4
\end{array} \right) , \quad \mathbf { y } _ { 3 } = \left( \begin{array} { r }
5
51
55
\end{array} \right)
\mathbf { P } = \left( \begin{array} { r r r }
1 & - 4 & 3
0 & 2 & 5
0 & 0 & - 7
\end{array} \right) .
\end{gathered}$$
- Show that \(\mathbf { y } _ { 1 } , \mathbf { y } _ { 2 } , \mathbf { y } _ { 3 }\) are linearly dependent.
- Find a basis for the linear space spanned by the vectors \(\mathbf { P y } _ { 1 } , \mathbf { P y } _ { 2 } , \mathbf { P y } _ { 3 }\).
4 Given that \(y = x \sin x\), find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\), simplifying your results as far as possible, and show that
$$\frac { \mathrm { d } ^ { 6 } y } { \mathrm {~d} x ^ { 6 } } = - x \sin x + 6 \cos x$$
Use induction to establish an expression for \(\frac { \mathrm { d } ^ { 2 n } y } { \mathrm {~d} x ^ { 2 n } }\), where \(n\) is a positive integer.
5 The integral \(I _ { n }\) is defined by
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \mathrm {~d} x$$
By considering \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \tan x \sec ^ { n } x \right)\), or otherwise, show that
$$( n + 1 ) I _ { n + 2 } = 2 ^ { \frac { 1 } { 2 } n } + n I _ { n }$$
Find the value of \(I _ { 6 }\).
6 Find the sum of the squares of the roots of the equation
$$x ^ { 3 } + x + 12 = 0$$
and deduce that only one of the roots is real.
The real root of the equation is denoted by \(\alpha\). Prove that \(- 3 < \alpha < - 2\), and hence prove that the modulus of each of the other roots lies between 2 and \(\sqrt { } 6\).
7 Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = \mathrm { e } ^ { - \alpha t }$$
where \(\alpha\) is a constant and \(\alpha \neq 2\).
Show that if \(\alpha < 2\) then, whatever the initial conditions, \(y \mathrm { e } ^ { \alpha t } \rightarrow \frac { 1 } { ( 2 - \alpha ) ^ { 2 } }\) as \(t \rightarrow \infty\).