4.10c Integrating factor: first order equations

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 \(y ^ { 2 } = \mathrm { f } ( x )\).

2. Find, in terms of \(k\), where \(k\) is a positive integer, the general solution of the differential equation

$$( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } + k y = x ^ { \frac { 1 } { 2 } } ( 1 + x ) ^ { 2 - k } , \quad x > 0$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(6)

1. Find the general solution of the differential equation

$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = \left( \cos ^ { 2 } x \right) \ln x , \quad 0 < x < \frac { \pi } { 2 }$$ Give your answer in the form \(y = \mathrm { f } ( x )\).

6. Obtain the general solution of the equation $$x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( x \cot x + 2 ) x y = 4 \sin x \quad 0 < x < \pi$$ Give your answer in the form \(y = \mathrm { f } ( x )\) (8)

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).

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) Determine the general solution of the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y \tan x = \mathrm { e } ^ { 4 x } \sec ^ { 3 } x$$ giving your answer in the form \(y = \mathrm { f } ( x )\) (b) Determine the particular solution for which \(y = 4\) at \(x = 0\)

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 \(x\).

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 substitution \(y = v x\) transforms the differential equation $$\frac { d y } { d x } = \frac { 3 x - 4 y } { 4 x + 3 y }$$ into the differential equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \frac { 3 v ^ { 2 } + 8 v - 3 } { 3 v + 4 }$$ (b) By solving differential equation (II), find a general solution of differential equation (I). (5)
(c) Given that \(y = 7\) at \(x = 1\), show that the particular solution of differential equation (I) can be written as $$( 3 y - x ) ( y + 3 x ) = 200$$ (5)(Total 14 marks)

6. (a) Find the general solution of the differential equation $$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( \sin x ) y = \cos ^ { 3 } x$$ (b) Show that, for \(0 \leq x \leq 2 \pi\), there are two points on the \(x\)-axis through which all the solution curves for this differential equation pass.
(c) Sketch the graph, for \(0 \leq x \leq 2 \pi\), of the particular solution for which \(y = 0\) at \(x = 0\).

6. (a) Using the substitution \(t = x ^ { 2 }\), or otherwise, find $$\int x ^ { 3 } \mathrm { e } ^ { - x ^ { 2 } } \mathrm {~d} x$$ (b) Find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x \mathrm { e } ^ { - x ^ { 2 } } , \quad x > 0$$

1. (a) Find the general solution of the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y = x$$ Given that \(y = 1\) at \(x = 0\),
(b) find the exact values of the coordinates of the minimum point of the particular solution curve,
(c) draw a sketch of this particular solution curve.

5. Find the general solution of the differential equation $$( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y = \frac { 1 } { x } , \quad x > 0 .$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(7)(Total 7 marks)

4. During an industrial process, the mass of salt, \(S \mathrm {~kg}\), dissolved in a liquid \(t\) minutes after the process begins is modelled by the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } + \frac { 2 S } { 120 - t } = \frac { 1 } { 4 } , \quad 0 \leq t < 120$$ Given that \(S = 6\) when \(t = 0\),

1. find \(S\) in terms of \(t\),
2. calculate the maximum mass of salt that the model predicts will be dissolved in the liquid at any one time during the process.
   (4)(Total 12 marks)

1. Obtain the general solution of the differential equation

$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = \cos x , \quad x > 0$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(Total 8 marks)

6. $$\frac { \mathrm { d } y } { \mathrm {~d} x } - y \tan x = 2 \sec ^ { 3 } x$$ Given that \(y = 3\) at \(x = 0\), find \(y\) in terms of \(x\) (Total 7 marks)

3. Find the general solution of the differential equation $$\sin x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y \cos x = \sin 2 x \sin x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

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).

3. Find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = \frac { \ln x } { x } , \quad x > 0$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

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

$$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 3 } + y ^ { 3 }$$ into the differential equation $$3 v ^ { 2 } x \frac { \mathrm {~d} v } { \mathrm {~d} x } = 1 - 2 v ^ { 3 }$$ (b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm { f } ( x )\). Given that \(y = 2\) at \(x = 1\),
(c) find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 1\)

1. (a) Find, in the form \(y = \mathrm { f } ( x )\), the general solution of the equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y \tan x = \sin 2 x , \quad 0 < x < \frac { \pi } { 2 }$$ Given that \(y = 2\) at \(x = \frac { \pi } { 3 }\) (b) find the value of \(y\) at \(x = \frac { \pi } { 6 }\), giving your answer in the form \(a + k \ln b\), where \(a\) and \(b\) are integers and \(k\) is rational.

4. Find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + ( 1 + x \cot x ) y = \sin x , \quad 0 < x < \pi$$ giving your answer in the form \(y = \mathrm { f } ( x )\).