Substitution to reduce order

10

1. By writing \(y = \sec ^ { 3 } \theta\) as \(\frac { 1 } { \cos ^ { 3 } \theta }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} \theta } = 3 \sin \theta \sec ^ { 4 } \theta\).
2. The variables \(x\) and \(\theta\) satisfy the differential equation $$\left( x ^ { 2 } + 9 \right) \sin \theta \frac { d \theta } { d x } = ( x + 3 ) \cos ^ { 4 } \theta$$ It is given that \(x = 3\) when \(\theta = \frac { 1 } { 3 } \pi\).
   Solve the differential equation to find the value of \(\cos \theta\) when \(x = 0\). Give your answer correct to 3 significant figures.
   If you use the following page to complete the answer to any question, the question number must be clearly shown.

2 Use the substitution \(z = x + y\) to find the solution of the differential equation $$\frac { d y } { d x } = \frac { 1 + 3 x + 3 y } { 3 x + 3 y - 1 }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\operatorname { aln } ( \mathrm { x } + \mathrm { y } ) + \mathrm { b } ( \mathrm { x } - \mathrm { y } ) + \mathrm { c } = 0\), where \(a , b\) and \(c\) are constants to be determined.

1. (a) Show that the transformation \(v = y - 2 x\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y x ( y - 4 x ) = 2 - 8 x ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} x } = - 2 x v ^ { 2 }$$ (b) Solve the differential equation (II) to determine \(v\) as a function of \(x\)
(c) Hence obtain the general solution of the differential equation (I).
(d) Sketch the solution curve that passes through the point \(( - 1 , - 1 )\). On your sketch show clearly the equation of any horizontal or vertical asymptotes.
You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.

4．The function \(\mathrm { h } ( x )\) has domain \(\mathbb { R }\) and range \(\mathrm { h } ( x ) > 0\) ，and satisfies $$\sqrt { \int \mathrm { h } ( x ) \mathrm { d } x } = \int \sqrt { \mathrm { h } ( x ) } \mathrm { d } x$$ （a）By substituting \(\mathrm { h } ( x ) = \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 }\) ，show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( y + c ) ,$$ where \(c\) is constant．
（b）Hence find a general expression for \(y\) in terms of \(x\) ．
（c）Given that \(\mathrm { h } ( 0 ) = 1\) ，find \(\mathrm { h } ( x )\) ．

5 The variables \(x\) and \(y\) are related by the differential equation $$x ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x y + x + 1 .$$

1. Use the substitution \(y = u - \frac { 1 } { x }\), where \(u\) is a function of \(x\), to show that the differential equation may be written as $$x ^ { 2 } \frac { \mathrm {~d} u } { \mathrm {~d} x } = u .$$
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = \mathrm { f } ( x )\).

3 The differential equation $$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y ^ { 3 } = \frac { \cos x } { x }$$ is to be solved for \(x > 0\). Use the substitution \(u = y ^ { 3 }\) to find the general solution for \(y\) in terms of \(x\).

2 Use the substitution \(u = y ^ { 2 }\) to find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = \frac { \mathrm { e } ^ { x } } { y }$$ for \(y\) in terms of \(x\).

3 The differential equation $$\frac { 2 } { y } - \frac { x } { y ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { 1 } { x ^ { 2 } }$$ is to be solved subject to the condition \(y = 1\) when \(x = 1\).

1. Show that \(y = \frac { 1 } { u }\) transforms the differential equation into $$x \frac { \mathrm {~d} u } { \mathrm {~d} x } + 2 u = \frac { 1 } { x ^ { 2 } } .$$
2. Find \(y\) in terms of \(x\).

Show that the substitution \(y = x z\) reduces the differential equation $$\frac { 1 } { x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { 6 } { x } - \frac { 2 } { x ^ { 2 } } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( \frac { 9 } { x } - \frac { 6 } { x ^ { 2 } } + \frac { 2 } { x ^ { 3 } } \right) y = 169 \sin 2 x$$ to the differential equation $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 9 z = 169 \sin 2 x$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , z = - 10\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = 5\).

10 It is given that \(x = t ^ { \frac { 1 } { 2 } }\), where \(x > 0\) and \(t > 0\), and \(y\) is a function of \(x\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$ reduces to the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
3. Find the general solution of ( \(*\) ), giving \(y\) in terms of \(x\).

The variable \(y\) depends on \(x\), and the variables \(x\) and \(t\) are related by \(x = \mathrm { e } ^ { t }\). Show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t } \quad \text { and } \quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } .$$

1. Given that \(y\) satisfies the differential equation $$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 16 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 25 y = 50 ( \ln x ) - 1$$ find a differential equation involving only \(t\) and \(y\).
2. Show that the complementary function of the differential equation in \(t\) and \(y\) may be written in the form $$R \mathrm { e } ^ { - \frac { 3 } { 2 } t } \sin ( 2 t + \phi )$$ where \(R\) and \(\phi\) are arbitrary constants.
3. Find a particular integral of the differential equation in \(t\) and \(y\).
4. Hence find the general solution of the differential equation in \(x\) and \(y\).

3

1. A differential equation is given by $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x$$ Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x }$$ transforms this differential equation into $$\frac { \mathrm { d } u } { \mathrm {~d} x } + \frac { 2 } { x } u = 3$$
2. Find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } + \frac { 2 } { x } u = 3$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

4

1. A differential equation is given by $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 }$$ Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x }$$ transforms this differential equation into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 1 } { x } u = 3 x$$
2. By using an integrating factor, find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 1 } { x } u = 3 x$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 }$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

8

1. Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
   1. \(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the substitution \(x = t ^ { 2 }\), where \(t \geqslant 0\), transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$ (2 marks)
3. Hence find the general solution of the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

6

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
   [0pt] [6 marks]
2. Show that the substitution \(u = x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\) transforms the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
3. Hence, given that \(x > 0\), find the general solution of the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ [2 marks]

1. A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.

The concentration, \(x\) parts per million (ppm), of the pollutant in the pond water \(t\) days after the chemical treatment was applied, is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 3 + \cosh t } { 3 x ^ { 2 } \cosh t } - \frac { 1 } { 3 } x \tanh t$$ When the chemical treatment was applied the concentration of pollutant was 3 ppm .

1. Use the iteration formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - y _ { n } \right) } { h }$$ once to estimate the concentration of the pollutant in the pond water 6 hours after the chemical treatment was applied.
2. Show that the transformation \(u = x ^ { 3 }\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} t } + u \tanh t = 1 + \frac { 3 } { \cosh t }$$
3. Determine the general solution of equation (II)
4. Hence find an equation for the concentration of pollutant in the pond water \(t\) days after the chemical treatment was applied.
5. Find the percentage error of the estimate found in part (a) compared to the value predicted by the model, stating if it is an overestimate or an underestimate.