4.10d Second order homogeneous: auxiliary equation method

3 The variables \(t\) and \(x\) are related by the differential equation $$\frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x } { d t } + x = t ^ { 2 } + 1$$

1. Find the general solution for \(x\) in terms of \(t\).
2. Deduce an approximate value of \(\frac { \mathrm { d } ^ { 2 } \mathrm { x } } { \mathrm { dt } ^ { 2 } }\) for large positive values of \(t\).

7 The variables \(x\) and \(y\) are related by the differential equation $$4 \frac { d ^ { 2 } y } { d x ^ { 2 } } - y = 3$$ It is given that, when \(x = 0 , y = - 3\) and \(\frac { \mathrm { dy } } { \mathrm { dx } } = 2\).

1. Find \(y\) in terms of \(x\).
2. Deduce the exact value of \(x\) for which \(y = 0\). Give your answer in logarithmic form.

6 Find the particular solution of the differential equation $$\frac { d ^ { 2 } x } { d t ^ { 2 } } - 12 \frac { d x } { d t } + 36 x = 37 \sin t$$ given that, when \(t = 0 , x = \frac { d x } { d t } = 0\).

2 The variables \(x\) and \(y\) are related by the differential equation $$6 \frac { d ^ { 2 } x } { d t ^ { 2 } } + 5 \frac { d x } { d t } + x = t ^ { 2 } + 10 t + 13$$

1. Find the general solution for \(x\) in terms of \(t\).
2. State an approximate solution for large positive values of \(t\).

6

1. Show that \(( \cosh x + \sinh x ) ^ { \frac { 1 } { 2 } } = \mathrm { e } ^ { \frac { 1 } { 2 } x }\).
2. Find the particular solution of the differential equation $$\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } + 3 y = 5 ( \cosh x + \sinh x ) ^ { \frac { 1 } { 2 } }$$ given that, when \(x = 0 , y = 1\) and \(\frac { d y } { d x } = \frac { 4 } { 3 }\).

5

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 10 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 25 x = 338 \sin t$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-10_2715_35_143_2012}
2. Show that, for large positive values of \(t\) and for any initial conditions, $$x \approx R \sin ( t - \phi ) ,$$ where the constants \(R\) and \(\phi\) are to be determined.

5 Find the particular solution of the differential equation $$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$ given that, when \(t = 0 , x = 12\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6\).
[0pt] [10] \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-10_2715_40_110_2007} \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-11_2726_35_97_20}

1 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = 7 - 2 t ^ { 2 }$$

8 The lifetime, in years, of an electrical component is the random variable \(T\), with probability density function f given by $$\mathrm { f } ( t ) = \begin{cases} A \mathrm { e } ^ { - \lambda t } & t \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$ where \(A\) and \(\lambda\) are positive constants.

1. Show that \(A = \lambda\). It is known that out of 100 randomly chosen components, 16 failed within the first year.
2. Find an estimate for the value of \(\lambda\), and hence find an estimate for the median value of \(T\).

1. The differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 8 \mathrm { e } ^ { - 3 t } \quad t \geqslant 0$$ describes the motion of a particle along the \(x\)-axis.

1. Determine the general solution of this differential equation. Given that the motion of the particle satisfies \(x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 }\) when \(t = 0\)
2. determine the particular solution for the motion of the particle. On the graph of the particular solution found in part (b), the first turning point for \(t > 0\) occurs at \(x = a\).
3. Determine, to 3 significant figures, the value of \(a\).
   [0pt] [Solutions relying entirely on calculator technology are not acceptable.]

8. (a) Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = 0 , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 0$$ (b) Hence find the general solution of the differential equation (I).

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).

6. (a) Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 8 y = 2 x ^ { 2 } + x$$ (b) Find the particular solution of this differential equation for which \(y = 1\) and $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 0 \text { when } x = 0$$

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

$$3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { 6 } { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \frac { 6 y } { x ^ { 2 } } + 3 y = x ^ { 2 } \quad x \neq 0$$ into the equation $$3 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 3 v = x$$ (b) Hence obtain the general solution of the differential equation (I), giving your answer in the form \(y = \mathrm { f } ( x )\)

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

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 48 x ^ { 2 } - 34$$ Given that \(y = 4\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 21\) at \(x = 0\) (b) determine the particular solution of the differential equation.
(c) Hence find the value of \(y\) at \(x = - 2\), giving your answer in the form \(p \mathrm { e } ^ { q } + r\) where \(p , q\) and \(r\) are integers to be determined.

14. (a) Find the value of \(\lambda\) for which \(\lambda x \cos 3 x\) is a particular integral of the differential equation $$\frac { d ^ { 2 } y } { d x ^ { 2 } } + 9 y = - 12 \sin 3 x$$ (b) Hence find the general solution of this differential equation.(4) The particular solution of the differential equation for which \(\boldsymbol { y } = \mathbf { 1 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathbf { 2 }\) at \(\boldsymbol { x } = \mathbf { 0 }\), is \(\boldsymbol { y } = \mathbf { g } ( \boldsymbol { x } )\).
(c) Find \(\mathrm { g } ( x )\).
(d) Sketch the graph of \(y = g ( x ) , 0 \leq x \leq \pi\).
(2) \section*{15.} \section*{Figure 1} Figure 1 shows a sketch of the cardioid \(C\) with equation \(r = a ( 1 + \cos \theta ) , - \pi < \theta \leq \pi\). Also shown are the tangents to \(C\) that are parallel and perpendicular to the initial line. These tangents form a rectangle WXYZ. \includegraphics[max width=\textwidth, alt={}, center]{141c7b1b-4236-4433-84af-04fa9baa3d96-5_407_782_315_1142}
(a) Find the area of the finite region, shaded in Fig. 1, bounded by the curve \(C\).
(b) Find the polar coordinates of the points \(A\) and \(B\) where \(W Z\) touches the curve \(C\).
(c) Hence find the length of \(W X\). Given that the length of \(\boldsymbol { W } \boldsymbol { Z }\) is \(\frac { 3 \sqrt { 3 } a } { 2 }\),
(d) find the area of the rectangle \(W X Y Z\). A heart-shape is modelled by the cardioid \(C\), where \(\boldsymbol { a } = \mathbf { 1 0 ~ c m }\). The heart shape is cut from the rectangular card WXYZ, shown in Fig. 1.
(e) Find a numerical value for the area of card wasted in making this heart shape.
8. A transformation \(T\) from the \(z\)-plane to the \(w\)-plane is defined by $$w = \frac { z + 1 } { i z - 1 } , \quad z \neq - i$$ where \(z = x + \mathrm { i } y , w = u + \mathrm { i } v\) and \(x , y , u\) and \(v\) are real. \(T\) transforms the circle \(| z | = 1\) in the \(z\)-plane onto a straight line \(L\) in the \(w\)-plane.
(a) Find an equation of \(L\) giving your answer in terms of \(u\) and \(v\).
(b) Show that \(T\) transforms the line \(\operatorname { Im } z = 0\) in the \(z\)-plane onto a circle \(C\) in the \(w\)-plane, giving the centre and radius of this circle.
(c) On a single Argand diagram sketch \(L\) and \(C\). Question: Solve $$x ^ { 5 } = - ( 9 \sqrt { 3 } ) i$$

3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in \(\mathrm { mg } l ^ { - 1 }\), at time \(t\) hours satisfies the differential equation $$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$

1. Show that the substitution \(\mathrm { y } = \frac { 1 } { x ^ { 2 } }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
2. Find the general solution of differential equation \(I\). Given that at time \(t = 0 , x = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\),
3. find an expression for \(x\) in terms of \(t\),
4. write down the maximum value of \(x\) as \(t\) varies.

7. For the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 x ( x + 3 )$$ find the solution for which at \(x = 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1\) and \(y = 1\).
(Total 12 marks)

8. $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 6 x = 2 \mathrm { e } ^ { - t }$$ Given that \(x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 2\) at \(t = 0\),

1. find \(x\) in terms of \(t\). The solution to part (a) is used to represent the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds, where \(t > 0 , P\) is \(x\) metres from the origin \(O\).
2. Show that the maximum distance between \(O\) and \(P\) is \(\frac { 2 \sqrt { } 3 } { 9 } \mathrm {~m}\) and justify that this
   distance is a maximum.

8. (a) Find the value of \(\lambda\) for which \(y = \lambda x \sin 5 x\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 25 y = 3 \cos 5 x$$ (b) Using your answer to part (a), find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 25 y = 3 \cos 5 x$$ Given that at \(x = 0 , y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 5\),
(c) find the particular solution of this differential equation, giving your solution in the form \(y = \mathrm { f } ( x )\).
(d) Sketch the curve with equation \(y = \mathrm { f } ( x )\) for \(0 \leqslant x \leqslant \pi\).

7. (a) Find the value of the constant \(\lambda\) for which \(y = \lambda x \mathrm { e } ^ { 2 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 y = 6 \mathrm { e } ^ { 2 x }$$ (b) Hence, or otherwise, find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 y = 6 \mathrm { e } ^ { 2 x }$$

3. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y - \sin x = 0$$ Given that \(y = \frac { 1 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 8 }\) at \(x = 0\), find a series expansion for \(y\) in terms of \(x\), up to and including the term in \(x ^ { 3 }\).

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

$$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 8 + 4 x ^ { 2 } \right) y = x ^ { 4 }$$ into the equation $$4 \frac { \mathrm {~d} ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 4 v = x$$ (b) Solve the differential equation (II) to find \(v\) as a function of \(x\).
(c) Hence state the general solution of the differential equation (I).

8. (a) Show that the substitution \(x = \mathrm { e } ^ { z }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 3 \ln x , \quad x > 0$$ into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} z ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} z } - 2 y = 3 z$$ (b) Find the general solution of the differential equation (II).
(c) Hence obtain the general solution of the differential equation (I) giving your answer in the form \(y = \mathrm { f } ( x )\). \(\square\)