Particular solution with initial conditions

8 Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 10 \mathrm { e } ^ { - 2 x }$$ given that \(y = 5\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = 0\).

8 Find the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 18 t ^ { 2 } + 6 t + 1$$ given that, when \(t = 0 , x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\).

7 Find the particular solution of the differential equation $$49 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 14 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 49 x + 735$$ given that when \(x = 0 , y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).

7 Find the particular solution of the differential equation $$10 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - x = t + 2$$ given that when \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\).

8 Find the particular solution of the differential equation $$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 50 \sin t$$ given that when \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\).

4 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 24 \mathrm { e } ^ { 2 x }$$ given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 9\) when \(x = 0\).

8 Find \(y\) in terms of \(t\), given that $$5 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 5 y = 15 + 12 t + 5 t ^ { 2 }$$ and that \(y = \frac { \mathrm { d } y } { \mathrm {~d} t } = 0\) when \(t = 0\).

9 Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 10 x = 2 \sin t - 3 \cos t$$ given that, when \(t = 0 , x = 3.3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.9\).

3

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5$$
2. Hence express \(y\) in terms of \(x\), given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) when \(x = 0\).

3 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 26 \mathrm { e } ^ { x }$$ given that \(y = 5\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 11\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).
(10 marks)

6

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 10 \mathrm { e } ^ { - 2 x } - 9$$ (10 marks)
2. Hence express \(y\) in terms of \(x\), given that \(y = 7\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0\) as \(x \rightarrow \infty\).

4 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 2 \mathrm { e } ^ { - x }$$ given that \(y \rightarrow 0\) as \(x \rightarrow \infty\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3\) when \(x = 0\).
[0pt] [10 marks] \includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-08_46_145_829_159}

7 Find the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 10 \mathrm { e } ^ { 4 x } + 8 \sin 2 x + 4 \cos 2 x$$ given that \(y = 2.5\) when \(x = 0\) and \(y = \frac { \pi } { 4 }\) when \(x = \frac { \pi } { 4 }\).
[0pt] [10 marks]

11 A particle is suspended in a resistive medium from one end of a light spring. The other end of the spring is attached to a point which is made to oscillate in a vertical line. The displacement of the particle may be modelled by the differential equation \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t\) where \(x\) is the displacement of the particle below the equilibrium position at time \(t\).
When \(t = 0\) the particle is stationary and its displacement is 2 .

1. Find the particular solution of the differential equation.
2. Write down an approximate equation for the displacement when \(t\) is large.

5 A capacitor is an electrical component which stores charge. The value of the charge stored by the capacitor, in suitable units, is denoted by \(Q\). The capacitor is placed in an electrical circuit. At any time \(t\) seconds, where \(t \geqslant 0 , Q\) can be modelled by the differential equation \(\frac { d ^ { 2 } Q } { d t ^ { 2 } } - 2 \frac { d Q } { d t } - 15 Q = 0\). Initially the charge is 100 units and it is given that \(Q\) tends to a finite limit as \(t\) tends to infinity.

1. Determine the charge on the capacitor when \(t = 0.5\).
2. Determine the finite limit of \(Q\) as \(t\) tends to infinity.

5. A particle \(P\) moves in a straight line. At time \(t\) seconds its displacement from a fixed point \(O\) on the line is \(x\) metres. The motion of \(P\) is modelled by the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 12 \cos 2 t - 6 \sin 2 t$$ When \(t = 0 , P\) is at rest at \(O\).

1. Find, in terms of \(t\), the displacement of \(P\) from \(O\).
2. Show that \(P\) comes to instantaneous rest when \(t = \frac { \pi } { 4 }\).
3. Find, in metres to 3 significant figures, the displacement of \(P\) from \(O\) when \(t = \frac { \pi } { 4 }\).
4. Find the approximate period of the motion for large values of \(t\). \section*{6.} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{618fdb9c-cc0b-4a80-a148-4311c908c94e-5_534_923_388_541}
   \end{figure} A small ball \(Q\) of mass \(2 m\) is at rest at the point \(B\) on a smooth horizontal plane. A second small ball \(P\) of mass \(m\) is moving on the plane with speed \(\frac { 13 } { 12 } u\) and collides with \(Q\). Both the balls are smooth, uniform and of the same radius. The point \(C\) is on a smooth vertical wall \(W\) which is at a distance \(d _ { 1 }\) from \(B\), and \(B C\) is perpendicular to \(W\). A second smooth vertical wall is perpendicular to \(W\) and at a distance \(d _ { 2 }\) from \(B\). Immediately before the collision occurs, the direction of motion of \(P\) makes an angle \(\alpha\) with \(B C\), as shown in Fig. 2, where tan \(\alpha = \frac { 5 } { 12 }\). The line of centres of \(P\) and \(Q\) is parallel to \(B C\). After the collision \(Q\) moves towards \(C\) with speed \(\frac { 3 } { 5 } u\).
5. Show that, after the collision, the velocity components of \(P\) parallel and perpendicular to \(C B\) are \(\frac { 1 } { 5 } u\) and \(\frac { 5 } { 12 } u\) respectively.
6. Find the coefficient of restitution between \(P\) and \(Q\).
7. Show that when \(Q\) reaches \(C , P\) is at a distance \(\frac { 4 } { 3 } d _ { 1 }\) from \(W\). For each collision between a ball and a wall the coefficient of restitution is \(\frac { 1 } { 2 }\).
   Given that the balls collide with each other again,
8. show that the time between the two collisions of the balls is \(\frac { 15 d _ { 1 } } { u }\),
9. find the ratio \(d _ { 1 } : d _ { 2 }\). \section*{END}

5. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2 m a k ^ { 2 }\), where \(k\) is a positive constant. The other end of the string is attached to a fixed point \(A\). At time \(t = 0 , P\) is released from rest from a point which is a distance \(2 a\) vertically below \(A\). When \(P\) is moving with speed \(v\), the air resistance has magnitude \(2 m k v\). At time \(t\), the extension of the string is \(x\).

1. Show that, while the string is taut, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = g$$ You are given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - k t } ( C \sin k t + D \cos k t ) + \frac { g } { 2 k ^ { 2 } } , \quad \text { where } C \text { and } D \text { are constants. }$$
2. Find the value of \(C\) and the value of \(D\). Assuming that the string remains taut,
3. find the value of \(t\) when \(P\) first comes to rest,
4. show that \(2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)\).

6. A particle \(P\) of mass 0.2 kg is suspended from a fixed point by a light elastic spring. The spring has natural length 0.8 m and modulus of elasticity 7 N . At time \(t = 0\) the particle is released from rest from a point 0.2 metres vertically below its equilibrium position. The motion of \(P\) is resisted by a force of magnitude \(2 v\) newtons, where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(P\). At time \(t\) seconds, \(P\) is \(x\) metres below its equilibrium position.

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 10 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 43.75 x = 0\)
2. Find \(x\) in terms of \(t\).
3. Find the value of \(t\) when \(P\) first comes to instantaneous rest.

5. A particle \(P\) of mass \(m\) is fixed to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2 m a n ^ { 2 }\). The other end of the string is attached to a fixed point \(O\). The particle \(P\) is released from rest at a point which is a distance \(2 a\) vertically below \(O\). The air resistance is modelled as having magnitude \(2 m n v\), where \(v\) is the speed of \(P\).

1. Show that, when the extension of the string is \(x\), $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 n \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 n ^ { 2 } x = g$$
2. Find \(x\) in terms of \(t\).

5. An elastic string spring of modulus \(2 m g\) and natural length \(l\) is fixed at one end. To the other end is attached a mass \(m\) which is allowed to hang in equilibrium. The mass is then pulled vertically downwards through a distance \(l\) and released from rest. The air resistance is modelled as having magnitude \(2 m \omega v\), where \(v\) is the speed of the particle and \(\omega = \sqrt { \frac { g } { l } }\). The particle is at distance \(x\) from its equilibrium position at time \(t\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 \omega ^ { 2 } x = 0\).
2. Find the general solution of this differential equation.
3. Hence find the period of the damped harmonic motion.

6. A particle \(P\) of mass 2 kg moves in the \(x - y\) plane. At time \(t\) seconds its position vector is \(\mathbf { r }\) metres. When \(t = 0\), the position vector of \(P\) is \(\mathbf { i }\) metres and the velocity of \(P\) is ( \(- \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The vector \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = \mathbf { 0 }$$

1. Find \(\mathbf { r }\) in terms of \(t\).
2. Show that the speed of \(P\) at time \(t\) is \(\mathrm { e } ^ { - t } \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Find, in terms of e, the loss of kinetic energy of \(P\) in the interval \(t = 0\) to \(t = 1\).

2. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the vector differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = \mathrm { e } ^ { 2 t } \mathbf { j }$$ When \(t = 0 , P\) has position vector \(( \mathbf { i } + \mathbf { j } ) \mathrm { m }\) and velocity \(2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find an expression for \(\mathbf { r }\) in terms of \(t\).

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

$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$ When \(t = 0\), the particle is at the origin and is moving with velocity \(( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
Find \(\mathbf { r }\) in terms of \(t\).

1. A particle \(P\) is moving in a plane. At time \(t\) seconds the position vector of \(P\) is \(\mathbf { r }\) metres and the velocity of \(P\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). When \(t = \frac { \pi } { 2 } , P\) is instantaneously at rest at the point with position vector \(( \mathbf { i } - \mathbf { j } ) \mathrm { m }\).

Given that \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = ( 3 \sin t ) \mathbf { i }$$ find \(\mathbf { v }\) in terms of \(t\).
(13)

2. [In this question, \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors in a horizontal plane and \(\mathbf { k }\) is a unit vector vertically upwards.] A particle of mass 2 kg moves under the action of a constant gravitational force \(- 19.6 \mathbf { k } \mathrm {~N}\). The particle is subject to a resistive force \(- \mathbf { v }\) newtons, where \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) is the velocity of the particle at time \(t\) seconds.

1. By writing down an equation of motion of the particle, show that \(\mathbf { v }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 0.5 \mathbf { v } = - 9.8 \mathbf { k }$$ When \(t = 0 , \mathbf { v } = ( 4 \mathbf { i } - 6 \mathbf { j } + 11.6 \mathbf { k } )\)
2. Find \(\mathbf { v }\) when \(t = \ln 4\)