Solve via substitution then back-substitute

7 It is given that \(x = t ^ { 3 } y\) and $$t ^ { 3 } \frac { d ^ { 2 } y } { d t ^ { 2 } } + \left( 4 t ^ { 3 } + 6 t ^ { 2 } \right) \frac { d y } { d t } + \left( 13 t ^ { 3 } + 12 t ^ { 2 } + 6 t \right) y = 61 e ^ { \frac { 1 } { 2 } t }$$

1. Show that $$\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 13 x = 61 e ^ { \frac { 1 } { 2 } t }$$
2. Find the general solution for \(y\) in terms of \(t\).

7 It is given that \(y = x ^ { 2 } w\) and $$x ^ { 2 } \frac { d ^ { 2 } w } { d x ^ { 2 } } + 4 x ( x + 1 ) \frac { d w } { d x } + \left( 5 x ^ { 2 } + 8 x + 2 \right) w = 5 x ^ { 2 } + 4 x + 2$$

1. Show that $$\frac { d ^ { 2 } y } { d x ^ { 2 } } + 4 \frac { d y } { d x } + 5 y = 5 x ^ { 2 } + 4 x + 2$$
2. Find the general solution for \(w\) in terms of \(x\).

8 It is given that \(\mathrm { y } = \operatorname { coshu }\), where \(u > 0\), and $$\sqrt { \cosh ^ { 2 } u - 1 } \left( \frac { d ^ { 2 } u } { d x ^ { 2 } } + \frac { d u } { d x } \right) + \cosh u \left( \frac { d u } { d x } \right) ^ { 2 } - 2 \cosh u = 4 e ^ { - x }$$

1. Show that $$\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } - 2 y = 4 e ^ { - x }$$
2. Find \(u\) in terms of \(x\), given that, when \(x = 0 , u = \ln 3\) and \(\frac { d u } { d x } = 3\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

8 It is given that \(\mathbf { v } = y ^ { 4 }\) and $$y ^ { 3 } \frac { d ^ { 2 } y } { d x ^ { 2 } } + 3 y ^ { 2 } \left( \frac { d y } { d x } \right) ^ { 2 } + y ^ { 3 } \frac { d y } { d x } + y ^ { 4 } = e ^ { - 2 x }$$

1. Show that $$\frac { d ^ { 2 } v } { d x ^ { 2 } } + \frac { d v } { d x } + 4 v = 4 e ^ { - 2 x }$$
2. Find \(y\) in terms of \(x\), given that, when \(x = 0 , y = 1\) and \(\frac { \mathrm { dy } } { \mathrm { dx } } = - \frac { 3 } { 8 }\).
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

8. (a) Show that the transformation \(x = \mathrm { e } ^ { u }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 8 y = 4 \ln x \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} u } - 8 y = 4 u$$ (b) Determine the general solution of differential equation (II), expressing \(y\) as a function of \(u\).
(c) Hence obtain the general solution of differential equation (I).

1. (a) Show that the transformation \(y = x v\) transforms the equation

$$3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 6 } { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 6 y } { x ^ { 2 } } + 3 y = x ^ { 2 } \quad x \neq 0$$ into the equation $$3 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 3 v = x$$ (b) Hence obtain the general solution of the differential equation (I), giving your answer in the form \(y = \mathrm { f } ( x )\)

1. (a) Given that \(t = \ln x\), where \(x > 0\), show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Hence show that the transformation \(t = \ln x\), where \(x > 0\), transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 y = 1 + 4 \ln x - 2 ( \ln x ) ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } - 2 y = 1 + 4 t - 2 t ^ { 2 }$$ (c) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(d) Hence determine the general solution of differential equation (I).

3. (a) Show that the transformation \(y = x v\) transforms the equation $$\begin{array} { l l } x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 2 + 9 x ^ { 2 } \right) y = x ^ { 5 } , \\ \text { into the equation } & \frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 9 v = x ^ { 2 } . \end{array}$$I (b) Solve the differential equation II to find \(v\) as a function of \(x\).
(c) Hence state the general solution of the differential equation I.
(1)(Total 12 marks)

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in \(\mathrm { mg } l ^ { - 1 }\), at time \(t\) hours satisfies the differential equation $$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$

1. Show that the substitution \(\mathrm { y } = \frac { 1 } { x ^ { 2 } }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
2. Find the general solution of differential equation \(I\). Given that at time \(t = 0 , x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\),
3. find an expression for \(x\) in terms of \(t\),
4. write down the maximum value of \(x\) as \(t\) varies.

1. (a) Show that the transformation \(y = x v\) transforms the equation

$$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 8 + 4 x ^ { 2 } \right) y = x ^ { 4 }$$ into the equation $$4 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 4 v = x$$ (b) Solve the differential equation (II) to find \(v\) as a function of \(x\).
(c) Hence state the general solution of the differential equation (I).

7. (a) Show that the transformation \(x = t ^ { 2 }\) transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 15 y = 15 x$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 15 y = 15 t ^ { 2 }$$ (b) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(c) Hence determine the general solution of differential equation (I). \includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-24_2258_53_308_1980}

11 Show that, with a suitable value of the constant \(\alpha\), the substitution \(y = x ^ { \alpha } w\) reduces the differential equation $$2 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( 3 x ^ { 2 } + 8 x \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( x ^ { 2 } + 6 x + 4 \right) y = \mathrm { f } ( x )$$ to $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = \mathrm { f } ( x )$$ Find the general solution for \(y\) in the case where \(\mathrm { f } ( x ) = 6 \sin 2 x + 7 \cos 2 x\).

The variables \(z\) and \(x\) are related by the differential equation $$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$ Use the substitution \(y = z ^ { 3 }\) to show that \(y\) and \(x\) are related by the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5 x + 2$$ Given that \(z = 1\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }\) when \(x = 0\), find \(z\) in terms of \(x\). Deduce that, for large positive values of \(x , z \approx x ^ { \frac { 1 } { 3 } }\).

7 The variables \(x\) and \(y\) are related by the differential equation $$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 3 } = 8 \mathrm { e } ^ { - x }$$ Given that \(v = y ^ { 3 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 15 v = 24 \mathrm { e } ^ { - x }$$ Hence find the general solution for \(y\) in terms of \(x\).

Show that the substitution \(v = \frac { 1 } { y }\) reduces the differential equation $$\frac { 2 } { y ^ { 3 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - \frac { 1 } { y ^ { 2 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 2 } { y ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 5 } { y } = 17 + 6 x - 5 x ^ { 2 }$$ to the differential equation $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 5 v = 17 + 6 x - 5 x ^ { 2 }$$ Hence find \(y\) in terms of \(x\), given that when \(x = 0 , y = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 1\).

10 It is given that \(t \neq 0\) and $$t \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 t x = 3 t ^ { 2 } + 1$$

1. Show that if \(y = t x\) then $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 3 t ^ { 2 } + 1$$
2. Find \(x\) in terms of \(t\), given that \(x = \frac { 1 } { 9 } \pi\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 }\) when \(t = \frac { 1 } { 3 } \pi\).

8 Given that $$2 y ^ { 3 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 12 y ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y ^ { 2 } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 17 y ^ { 4 } = 13 \mathrm { e } ^ { - 4 x }$$ and that \(v = y ^ { 4 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 34 v = 26 \mathrm { e } ^ { - 4 x }$$ Hence find the general solution for \(y\) in terms of \(x\).

11 It is given that \(x \neq 0\) and $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 x y = 8 x ^ { 2 } + 16$$ Show that if \(z = x y\) then $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 z = 8 x ^ { 2 } + 16$$ Find \(y\) in terms of \(x\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2\) when \(x = \frac { 1 } { 2 } \pi\).

Given that $$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y ^ { 3 } = 25 \mathrm { e } ^ { - 2 x }$$ and that \(v = y ^ { 3 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 9 v = 75 \mathrm { e } ^ { - 2 x }$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\). \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

9 Given that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + ( 2 - 3 x ) y = 10 \mathrm { e } ^ { 2 x }$$ and that \(v = x y\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 3 v = 10 \mathrm { e } ^ { 2 x }$$ Find the general solution for \(y\) in terms of \(x\).

It is given that \(w = \cos y\) and $$\tan y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 2 \tan y \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 + \mathrm { e } ^ { - 2 x } \sec y$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = - \mathrm { e } ^ { - 2 x }$$
2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = \frac { 1 } { 3 } \pi\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 3 } }\). [10]

Show that the substitution \(y = \frac { 1 } { w }\) reduces the differential equation $$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 2 } = \left( 5 x ^ { 2 } + 4 x + 2 \right) y ^ { 3 }$$ to $$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + 5 w = - 5 x ^ { 2 } - 4 x - 2$$ Find the general solution for \(w\) in terms of \(x\). Find a function f such that \(\lim _ { x \rightarrow \infty } \left( \frac { y } { \mathrm { f } ( x ) } \right) = 1\).

Given that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x ( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 \left( 1 + 4 x + 2 x ^ { 2 } \right) y = 8 x ^ { 2 }$$ and that \(x ^ { 2 } y = z\), show that $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 4 z = 8 x ^ { 2 }$$ Find the general solution for \(y\) in terms of \(x\). Describe the behaviour of \(y\) as \(x \rightarrow \infty\).

7 It is given that, for \(x \neq 0 , y\) satisfies the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$

1. Show that the substitution \(u = x y\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} u } { \mathrm {~d} x } + 9 u = 18 x$$
2. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (8 marks)