Bernoulli equation

4. (a) Show that the substitution \(y ^ { 2 } = \frac { 1 } { z }\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y = 3 x y ^ { 3 } \quad y \neq 0$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 z = - 6 x$$ (b) Obtain the general solution of differential equation (II).
(c) Hence obtain the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\)

1. (a) Show that the transformation \(y = \frac { 1 } { z }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 2 y ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { z } { x } = - \frac { 2 } { x ^ { 2 } }$$ (b) Solve differential equation (II) to determine \(z\) in terms of \(x\).
(c) Hence determine the particular solution of differential equation (I) for which \(y = - \frac { 3 } { 8 }\) at \(x = 3\) Give your answer in the form \(y = \mathrm { f } ( x )\).

1. (a) For all the values of \(x\) where the identity is defined, prove that

$$\cot 2 x + \tan x \equiv \operatorname { cosec } 2 x$$ (b) Show that the substitution \(y ^ { 2 } = w \sin 2 x\), where \(w\) is a function of \(x\), transforms the differential equation $$y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } \tan x = \sin x \quad 0 < x < \frac { \pi } { 2 }$$ into the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} x } + 2 w \operatorname { cosec } 2 x = \sec x \quad 0 < x < \frac { \pi } { 2 }$$ (c) By solving differential equation (II), determine a general solution of differential equation (I) in the form \(y ^ { 2 } = \mathrm { f } ( x )\), where \(\mathrm { f } ( x )\) is a function in terms of \(\cos x\) $$\text { [You may use without proof } \left. \int \operatorname { cosec } 2 x \mathrm {~d} x = \frac { 1 } { 2 } \ln | \tan x | \text { (+ constant) } \right]$$

1. (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 x y = x \mathrm { e } ^ { - x ^ { 2 } } y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 x z = - 2 x \mathrm { e } ^ { - x ^ { 2 } }$$ (b) Solve differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence find the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

8. (a) Show that the substitution \(v = y ^ { - 2 }\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 6 x y = 3 x \mathrm { e } ^ { x ^ { 2 } } y ^ { 3 } \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} x } - 12 v x = - 6 x \mathrm { e } ^ { x ^ { 2 } } \quad x > 0$$ (b) Hence find the general solution of the differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation

$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y + 4 x ^ { 2 } y ^ { 3 } \ln x = 0 \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - \frac { 2 z } { x } = 8 x \ln x \quad x > 0$$ (b) By solving differential equation (II), determine the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\)

1. (a) Show that the substitution \(y ^ { 2 } = \frac { 1 } { t }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + y = x y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } t } { \mathrm {~d} x } - 2 t = - 2 x$$ (b) Solve differential equation (II) and determine \(y ^ { 2 }\) in terms of \(x\).

7. (a) Show that the transformation \(z = y ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$ (b) Solve the differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence obtain the general solution of the differential equation (I).

7. (a) Show that the substitution \(v = y ^ { - 3 }\) transforms the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 2 x ^ { 4 } y ^ { 4 }$$ into the differential equation $$\begin{aligned} & \frac { \mathrm { d } v } { \mathrm {~d} x } - \frac { 3 v } { x } = - 6 x ^ { 3 } \\ & \text { ration (II), find a general solution of differential equation (I) } \end{aligned}$$ in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

1. (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 x y = x \mathrm { e } ^ { - x ^ { 2 } } y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 x z = - 2 x \mathrm { e } ^ { - x ^ { 2 } }$$ (b) Solve differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence find the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. (a) Show that the transformation \(z = y ^ { \frac { 1 } { 2 } }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$ (b) Solve the differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence obtain the general solution of the differential equation (I).
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5 The substitution \(y = u ^ { k }\), where \(k\) is an integer, is to be used to solve the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } y ^ { 2 }$$ by changing it into an equation (B) in the variables \(u\) and \(x\).

1. Show that equation (B) may be written in the form $$\frac { \mathrm { d } u } { \mathrm {~d} x } + \frac { 3 } { k x } u = \frac { 1 } { k } x u ^ { k + 1 }$$
2. Write down the value of \(k\) for which the integrating factor method may be used to solve equation (B).
3. Using this value of \(k\), solve equation (B) and hence find the general solution of equation (A), giving your answer in the form \(y = \mathrm { f } ( x )\).
   (a) The set of polynomials \(\{ a x + b \}\), where \(a , b \in \mathbb { R }\), is denoted by \(P\). Assuming that the associativity property holds, prove that \(P\), under addition, is a group.
   (b) The set of polynomials \(\{ a x + b \}\), where \(a , b \in \{ 0,1,2 \}\), is denoted by \(Q\). It is given that \(Q\), under addition modulo 3 , is a group, denoted by \(( Q , + ( \bmod 3 ) )\).
4. State the order of the group.
5. Write down the inverse of the element \(2 x + 1\).
6. \(\mathrm { q } ( x ) = a x + b\) is any element of \(Q\) other than the identity. Find the order of \(\mathrm { q } ( x )\) and hence determine whether \(( Q , + ( \bmod 3 ) )\) is a cyclic group.