Substitution reducing to first order linear ODE

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 $$\sqrt { \int \mathrm { h } ( x ) \mathrm { d } x } = \int \sqrt { \mathrm { h } ( x ) } \mathrm { d } x$$

1. 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.
2. Hence find a general expression for \(y\) in terms of \(x\) .
3. 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 \(x\).

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 \(y = \mathrm { g } ( x )\).

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.

The variables \(x\) and \(y\) are related by the differential equation $$x^3 \frac{dy}{dx} = xy + x + 1. \qquad (A)$$

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 $$x^2 \frac{du}{dx} = u.$$ [4]
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = f(x)\). [5]