6 A differential equation is given by
$$\left( x ^ { 3 } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 3 x ^ { 2 } \frac { d y } { d x } = 2 - 4 x ^ { 3 }$$
- Show that the substitution
$$u = \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 x$$
transforms this differential equation into
$$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } u } { \mathrm {~d} x } = 3 x ^ { 2 } u$$
(4 marks)
- Hence find the general solution of the differential equation
$$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 - 4 x ^ { 3 }$$
giving your answer in the form \(y = \mathrm { f } ( x )\).
\(7 \quad\) The curve \(C _ { 1 }\) is defined by \(r = 2 \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\).
The curve \(C _ { 2 }\) is defined by \(r = \tan \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }\). - Find a cartesian equation of \(C _ { 1 }\).
- Prove that the curves \(C _ { 1 }\) and \(C _ { 2 }\) meet at the pole \(O\) and at one other point, \(P\), in the given domain. State the polar coordinates of \(P\).
- The point \(A\) is the point on the curve \(C _ { 1 }\) at which \(\theta = \frac { \pi } { 4 }\).
The point \(B\) is the point on the curve \(C _ { 2 }\) at which \(\theta = \frac { \pi } { 4 }\).
Determine which of the points \(A\) or \(B\) is further away from the pole \(O\), justifying your answer.
- Show that the area of the region bounded by the arc \(O P\) of \(C _ { 1 }\) and the arc \(O P\) of \(C _ { 2 }\) is \(a \pi + b \sqrt { 3 }\), where \(a\) and \(b\) are rational numbers.