AQA FP3 2008 January — Question 8

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2008
SessionJanuary
TopicSecond order differential equations
8
  1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that:
    1. \(x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
    2. \(\quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }\).
  2. Hence find the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 0$$
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