First order differential equations (integrating factor)
8
Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
\(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
\(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
Hence show that the substitution \(x = t ^ { 2 }\), where \(t \geqslant 0\), transforms the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$
(2 marks)
Hence find the general solution of the differential equation
$$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$
giving your answer in the form \(y = \mathrm { g } ( x )\).