Substitution changing independent variable

Show that the substitution \(y = x z\) reduces the differential equation $$\frac { 1 } { x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { 6 } { x } - \frac { 2 } { x ^ { 2 } } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( \frac { 9 } { x } - \frac { 6 } { x ^ { 2 } } + \frac { 2 } { x ^ { 3 } } \right) y = 169 \sin 2 x$$ to the differential equation $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 9 z = 169 \sin 2 x$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , z = - 10\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = 5\).

10 It is given that \(x = t ^ { \frac { 1 } { 2 } }\), where \(x > 0\) and \(t > 0\), and \(y\) is a function of \(x\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$ reduces to the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
3. Find the general solution of ( \(*\) ), giving \(y\) in terms of \(x\).

The variable \(y\) depends on \(x\), and the variables \(x\) and \(t\) are related by \(x = \mathrm { e } ^ { t }\). Show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t } \quad \text { and } \quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } .$$

1. Given that \(y\) satisfies the differential equation $$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 16 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 25 y = 50 ( \ln x ) - 1$$ find a differential equation involving only \(t\) and \(y\).
2. Show that the complementary function of the differential equation in \(t\) and \(y\) may be written in the form $$R \mathrm { e } ^ { - \frac { 3 } { 2 } t } \sin ( 2 t + \phi )$$ where \(R\) and \(\phi\) are arbitrary constants.
3. Find a particular integral of the differential equation in \(t\) and \(y\).
4. Hence find the general solution of the differential equation in \(x\) and \(y\).

8

1. Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
   1. \(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
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)
3. 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 \(y = \mathrm { g } ( x )\).