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). $$\left[ \begin{array} { l } \text { Leave } \\ \text { blank } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \end{array} \right.$$

9

1. Show that the substitution \(u = \frac { 1 } { y ^ { 3 } }\) transforms the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\) into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x$$
2. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\), given that \(y = \frac { 1 } { 2 }\) when \(x = 0\). Give your answer in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

9

1. Show that the substitution \(u = \frac { 1 } { y ^ { 3 } }\) transforms the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\) into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x .$$
2. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\), given that \(y = \frac { 1 } { 2 }\) when \(x = 0\). Give your answer in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

9

1. Show that the substitution \(u = \frac { 1 } { y ^ { 3 } }\) transforms the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\) into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - 3 u = - 9 x$$
2. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + y = 3 x y ^ { 4 }\), given that \(y = \frac { 1 } { 2 }\) when \(x = 0\). Give your answer in the form \(y ^ { 3 } = \mathrm { f } ( x )\).

The substitution \(y = u^k\), where \(k\) is an integer, is to be used to solve the differential equation $$x \frac{dy}{dx} + 3y = x^2 y^2 \qquad (A)$$ by changing it into an equation (B) in the variables \(u\) and \(x\).

1. Show that equation (B) may be written in the form $$\frac{du}{dx} + \frac{3}{kx} u = \frac{1}{k} x u^{k+1}.$$ [4]
2. Write down the value of \(k\) for which the integrating factor method may be used to solve equation (B). [1]
3. Using this value of \(k\), solve equation (B) and hence find the general solution of equation (A), giving your answer in the form \(y = f(x)\). [4]