4.10c Integrating factor: first order equations

4 The differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { 1 - x ^ { 2 } } y = ( 1 - x ) ^ { \frac { 1 } { 2 } } , \quad \text { where } | x | < 1$$ can be solved by the integrating factor method.

1. Use an appropriate result given in the List of Formulae (MF1) to show that the integrating factor can be written as \(\left( \frac { 1 + x } { 1 - x } \right) ^ { \frac { 1 } { 2 } }\).
2. Hence find the solution of the differential equation for which \(y = 2\) when \(x = 0\), giving your answer in the form \(y = \mathrm { f } ( x )\).

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 \(x\).

7 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?

8 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
4. What does this solution indicate about the long-term height of the tree?
5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?

7 The variables \(x\) and \(y\) are related by the differential equation $$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 3 } = 8 \mathrm { e } ^ { - x }$$ Given that \(v = y ^ { 3 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 15 v = 24 \mathrm { e } ^ { - x }$$ Hence find the general solution for \(y\) in terms of \(x\).

9 Show that if \(y\) depends on \(x\) and \(x = \mathrm { e } ^ { u }\) then $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} u } .$$ Given that \(y\) satisfies the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 30 x ^ { 2 }$$ use the substitution \(x = \mathrm { e } ^ { u }\) to show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 3 y = 30 \mathrm { e } ^ { 2 u }$$ Hence find the general solution for \(y\) in terms of \(x\).

9 Given that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + ( 2 - 3 x ) y = 10 \mathrm { e } ^ { 2 x }$$ and that \(v = x y\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 3 v = 10 \mathrm { e } ^ { 2 x }$$ Find the general solution for \(y\) in terms of \(x\).

It is given that \(w = \cos y\) and $$\tan y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 2 \tan y \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 + \mathrm { e } ^ { - 2 x } \sec y$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = - \mathrm { e } ^ { - 2 x }$$
2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = \frac { 1 } { 3 } \pi\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 3 } }\). [10]

5 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 x } { x ^ { 2 } + 1 } y = x$$ given that \(y = 1\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).

2

1. Show that \(\frac { 1 } { x ^ { 2 } }\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 2 } { x } y = x$$
2. Hence find the general solution of this differential equation, giving your answer in the form \(y = \mathrm { f } ( x )\).

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

2

1. Find the values of the constants \(p\) and \(q\) for which \(p \sin x + q \cos x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 5 y = 13 \cos x$$
2. Hence find the general solution of this differential equation.

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 2 } { x } y = 2 x ^ { 3 } \mathrm { e } ^ { 2 x }$$ given that \(y = \mathrm { e } ^ { 4 }\) when \(x = 2\). Give your answer in the form \(y = \mathrm { f } ( x )\).

4

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 } { x } y = \ln x$$
2. Hence, given that \(y \rightarrow 0\) as \(x \rightarrow 0\), find the value of \(y\) when \(x = 1\).

5

1. Show that \(\tan x\) is an integrating factor for the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { \sec ^ { 2 } x } { \tan x } y = \tan x$$ (2 marks)
2. Hence solve this differential equation, given that \(y = 3\) when \(x = \frac { \pi } { 4 }\).
   (6 marks)

3

1. Show that \(\sin x\) is an integrating factor for the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \cot x ) y = 2 \cos x$$
2. Solve this differential equation, given that \(y = 2\) when \(x = \frac { \pi } { 2 }\).

6

1. Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y$$ transforms the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$ (4 marks)
2. By using an integrating factor, or otherwise, find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

2

1. Find the values of the constants \(a , b , c\) and \(d\) for which \(a + b x + c \sin x + d \cos x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 10 \sin x - 3 x$$ (4 marks)
2. Hence find the general solution of this differential equation.

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

2 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - y \tan x = 2 \sin x$$ given that \(y = 2\) when \(x = 0\).
(9 marks)

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 3 } { x } y = \left( x ^ { 4 } + 3 \right) ^ { \frac { 3 } { 2 } }$$ given that \(y = \frac { 1 } { 5 }\) when \(x = 1\).
(9 marks)

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \cot x ) y = \sin 2 x , \quad 0 < x < \frac { \pi } { 2 }$$ given that \(y = \frac { 1 } { 2 }\) when \(x = \frac { \pi } { 6 }\).
(10 marks)

4

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 } { 2 x + 1 } y = 4 ( 2 x + 1 ) ^ { 5 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
2. The gradient of a curve at any point \(( x , y )\) on the curve is given by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 ( 2 x + 1 ) ^ { 5 } - \frac { 4 } { 2 x + 1 } y$$ The point whose \(x\)-coordinate is zero is a stationary point of the curve. Using your answer to part (a), find the equation of the curve.

5

1. Differentiate \(\ln ( \ln x )\) with respect to \(x\).
2. 1. Show that \(\ln x\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { x \ln x } y = 9 x ^ { 2 } , \quad x > 1$$
   2. Hence find the solution of this differential equation, given that \(y = 4 \mathrm { e } ^ { 3 }\) when \(x = \mathrm { e }\).
      (6 marks)

2

1. Find the values of the constants \(a\), \(b\) and \(c\) for which \(a + b \sin 2 x + c \cos 2 x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 4 y = 20 - 20 \cos 2 x$$ [4 marks]
2. Hence find the solution of this differential equation, given that \(y = 4\) when \(x = 0\).
   [0pt] [4 marks]
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