Standard linear first order

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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2 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \tan ^ { 3 } x \sec x$$ given that \(y = 2\) when \(x = \frac { \pi } { 3 }\).
[0pt] [9 marks]

6

1. Obtain the general solution of the differential equation $$\tan x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = \sin x \tan x$$ where \(0 < x < \frac { \pi } { 2 }\)
   [0pt] [5 marks]
   6
2. Hence find the particular solution of this differential equation, given that \(y = \frac { 1 } { 2 \sqrt { 2 } }\)
   when \(x = \frac { \pi } { 4 }\)
   [0pt] [2 marks]

3. The position vector \(\mathbf { r }\) of a particle \(P\) at time \(t\) satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 4 \mathbf { i }$$ Given that the position vector of \(P\) at time \(t = 0\) is \(2 \mathbf { j }\), find the position vector of \(P\) at time \(t\).
(Total 6 marks)

2. With respect to a fixed origin \(O\), the position vector, \(\mathbf { r }\) metres, of a particle \(P\) at time \(t\) seconds satisfies $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \mathbf { r } = ( \mathbf { i } - \mathbf { j } ) \mathrm { e } ^ { - 2 t } .$$ Given that \(P\) is at \(O\) when \(t = 0\), find

1. \(\mathbf { r }\) in terms of \(t\),
2. a cartesian equation of the path of \(P\).

2. At time \(t\) seconds the position vector of a particle \(P\), relative to a fixed origin \(O\), is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 3 \mathrm { e } ^ { - t } \mathbf { j }$$ Given that \(\mathbf { r } = 2 \mathbf { i } - \mathbf { j }\) when \(t = 0\), find \(\mathbf { r }\) in terms of \(t\).
(Total 7 marks)

2. The velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$ The position vector of \(P\) at time \(t\) seconds is \(\mathbf { r }\) metres.
Given that at \(t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )\) and \(\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )\), find \(\mathbf { r }\) at time \(t\) seconds.

1. At time \(t = 0\), the position vector of a particle \(P\) is \(- 3 \mathbf { j } \mathrm {~m}\). At time \(t\) seconds, the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that

$$\mathbf { v } - 2 \mathbf { r } = 4 \mathrm { e } ^ { t } \mathbf { j }$$ find the time when \(P\) passes through the origin.

1. A particle \(P\) moves in a plane such that its position vector \(\mathbf { r }\) metres at time \(t\) seconds \(( t > 0 )\) satisfies the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - \frac { 2 } { t } \mathbf { r } = 4 \mathbf { i }$$ When \(t = 1\), the particle is at the point with position vector \(( \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).

1. Solve the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - 2 \mathbf { r } = \mathbf { 0 }$$ given that when \(t = 0 , \mathbf { r } . \mathbf { j } = 0\) and \(\mathbf { r } \times \mathbf { j } = \mathbf { i } + \mathbf { k }\).

1. A particle moves in a plane so that its position vector \(\mathbf { r }\) metres at time \(t\) seconds satisfies the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + ( \tan t ) \mathbf { r } = \left( \cos ^ { 2 } t \right) \mathbf { i } - ( 3 \cos t ) \mathbf { j } , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ When \(t = 0\), the particle is at the point with position vector \(4 \mathbf { j } \mathrm {~m}\). Find \(\mathbf { r }\) in terms of \(t\).

2. A particle \(P\) moves so that its position vector, \(\mathbf { r }\) metres, at time \(t\) seconds, where \(0 \leqslant t < \frac { \pi } { 2 }\), satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } - ( \tan t ) \mathbf { r } = ( \sin t ) \mathbf { i }$$ When \(t = 0 , \mathbf { r } = - \frac { 1 } { 2 } \mathbf { i }\).
Find \(\mathbf { r }\) in terms of \(t\).

3. A particle \(P\) moves in the \(x y\)-plane in such a way that its position vector \(\mathbf { r }\) metres at time \(t\) seconds, where \(0 \leqslant t < \pi\), satisfies the differential equation $$\sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \sec ^ { 3 } \left( \frac { 1 } { 2 } t \right) \sin \left( \frac { 1 } { 2 } t \right) \mathbf { r } = \sin \left( \frac { 1 } { 2 } t \right) \mathbf { i } + \sec ^ { 2 } \left( \frac { 1 } { 2 } t \right) \mathbf { j }$$ When \(t = 0\), the particle is at the point with position vector \(( - \mathbf { i } + \mathbf { j } ) \mathrm { m }\).
Find \(\mathbf { r }\) in terms of \(t\).

12 Solve the differential equation \(\left( 4 - x ^ { 2 } \right) \frac { d y } { d x } - x y = 1\), given that \(y = 1\) when \(x = 0\), giving your answer in the form \(y = \mathrm { f } ( x )\).

11 Solve the differential equation \(\cosh x \frac { d y } { d x } - 2 y \sinh x = \cosh x\), given that \(y = 1\) when \(x = 0\).

10. Given the differential equation $$\sec x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \operatorname { cosec } x = 2$$ and \(x = \frac { \pi } { 2 }\) when \(y = 3\), find the value of \(y\) when \(x = \frac { \pi } { 4 }\).

9. Consider the differential equation $$( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 5 y = ( x + 1 ) ^ { 2 } , \quad x > - 1$$ Given that \(y = \frac { 1 } { 4 }\) when \(x = 1\), find the value of \(y\) when \(x = 0\).

1. Given the differential equation

$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = 4 \cos ^ { 3 } x \sin x + 5$$ and \(y = 3 \sqrt { 2 }\) when \(x = \frac { \pi } { 4 }\), find an equation for \(y\) in terms of \(x\).
□

10. Consider the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y \tan x = \sin x , \quad 0 < x < \frac { \pi } { 2 }$$

1. Find an integrating factor for this differential equation.
2. Solve the differential equation given that \(y = 0\) when \(x = \frac { \pi } { 4 }\), giving your answer in the form \(y = f ( x )\).

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

$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = \mathrm { e } ^ { 2 x } \cos ^ { 2 } x$$ giving your answer in the form \(y = \mathrm { f } ( x )\) Given that \(y = 3\) when \(x = 0\)
(b) determine the smallest positive value of \(x\) for which \(y = 0\)

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 \(x = 0\).
[0pt] [P4 June 2002 Qn 6]

3

1. Show that \(y = x ^ { 3 } - x\) is a particular integral of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$
2. By differentiating \(\left( x ^ { 2 } - 1 \right) y = c\) implicitly, where \(y\) is a function of \(x\) and \(c\) is a constant, show that \(y = \frac { c } { x ^ { 2 } - 1 }\) is a solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 0$$
3. Hence find the general solution of $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 x y } { x ^ { 2 } - 1 } = 5 x ^ { 2 } - 1$$

5

1. The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x \ln x + \frac { y } { x }$$ and $$y ( 1 ) = 1$$
   1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\).
   2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a)(i) to obtain an approximation to \(y ( 1.2 )\), giving your answer to three decimal places.
2. 1. Show that \(\frac { 1 } { x }\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 1 } { x } y = x \ln x$$
   2. Solve this differential equation, given that \(y = 1\) when \(x = 1\).
   3. Calculate the value of \(y\) when \(x = 1.2\), giving your answer to three decimal places.

3

1. Show that \(x ^ { 2 }\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 } { x } y = 3 \left( x ^ { 3 } + 1 \right) ^ { \frac { 1 } { 2 } }$$
2. Solve this differential equation, given that \(y = 1\) when \(x = 2\).