Euler-Cauchy equations via exponential substitution

8. (a) Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = 0 , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 0$$ (b) Hence find the general solution of the differential equation (I).

8. (a) Show that the substitution \(x = \mathrm { e } ^ { z }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 3 \ln x , \quad x > 0$$ into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} z ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} z } - 2 y = 3 z$$ (b) Find the general solution of the differential equation (II).
(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form \(y = \mathrm { f } ( x )\). \(\square\)

1. (a) Show that the transformation \(x = \mathrm { e } ^ { u }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 7 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 2 \ln x , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 16 y = 2 u$$ (b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(u\).
(c) Hence obtain the general solution of the differential equation (I).

7. (a) Show that the substitution \(x = e ^ { u }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = - x ^ { - 2 } , \quad x > 0$$ into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 2 y = - \mathrm { e } ^ { - 2 u }$$ (b) Find the general solution of the differential equation (II).
(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form \(y = \mathrm { f } ( x )\)

7. (a) Given that \(x = e ^ { t }\), show that

1. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \frac { \mathrm {~d} y } { \mathrm {~d} t }$$
2. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Use you answers to part (a) to show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = \mathrm { e } ^ { 3 t }$$ (c) Hence find the general solution of $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$

1. (a) Show that the transformation \(x = \mathrm { e } ^ { t }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { 2 t }$$ (b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(t\).
(c) Hence find the general solution of the differential equation (I).

9 Show that if \(y\) depends on \(x\) and \(x = \mathrm { e } ^ { u }\) then $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} u } .$$ Given that \(y\) satisfies the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 30 x ^ { 2 }$$ use the substitution \(x = \mathrm { e } ^ { u }\) to show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 3 y = 30 \mathrm { e } ^ { 2 u }$$ Hence find the general solution for \(y\) in terms of \(x\).

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 $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 0$$

7

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
3. Find the general solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10\).
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 8\) when \(x = 1\).

8

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
3. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x\), given that \(y = \frac { 3 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }\) when \(x = 1\).
   (5 marks)

7

1. Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$\text { into } \quad \begin{aligned} x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x ) \\ \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t \end{aligned}$$ (7 marks)
2. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$ (11 marks)
3. Write down the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$