Particular solution with initial conditions

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 \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\).

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 \(P\) 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 \(t\).

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 \(t = \ln 4\)

3. At time \(t\) seconds, the position vector \(\mathbf { r }\) metres of a particle \(P\), relative to a fixed origin \(O\), satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 3 \mathbf { r } = \mathbf { 0 }$$ At time \(t = 0 , P\) is at the point with position vector \(2 \mathbf { i } \mathrm {~m}\) and is moving with velocity \(2 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Find the position vector of \(P\) when \(t = \ln 2\).
(10 marks)

14

1. Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } - 2 y = 12 e ^ { - x }\). You are given that \(y\) tends to zero as \(x\) tends to infinity, and that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 0\) when \(x = 0\).
2. Find the exact value of \(x\) for which \(y = 0\).

1. An engineer is investigating the motion of a sprung diving board at a swimming pool.

Let \(E\) be the position of the end of the diving board when it is at rest in its equilibrium position and when there is no diver standing on the diving board.
A diver jumps from the diving board.
The vertical displacement, \(h \mathrm {~cm}\), of the end of the diving board above \(E\) is modelled by the differential equation $$4 \frac { \mathrm {~d} ^ { 2 } h } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} h } { \mathrm {~d} t } + 37 h = 0$$ where \(t\) seconds is the time after the diver jumps.

1. Find a general solution of the differential equation. When \(t = 0\), the end of the diving board is 20 cm below \(E\) and is moving upwards with a speed of \(55 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).
2. Find, according to the model, the maximum vertical displacement of the end of the diving board above \(E\).
3. Comment on the suitability of the model for large values of \(t\).

1. A scientist is investigating the concentration of antibodies in the bloodstream of a patient following a vaccination.
   The concentration of antibodies, \(x\), measured in micrograms ( \(\mu \mathrm { g }\) ) per millilitre ( ml ) of blood, is modelled by the differential equation

$$100 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 60 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 26$$ where \(t\) is the number of weeks since the vaccination was given.

1. Find a general solution of the differential equation. Initially,
   • there are no antibodies in the bloodstream of the patient
   • the concentration of antibodies is estimated to be increasing at \(10 \mu \mathrm {~g} / \mathrm { ml }\) per week
   • Find, according to the model, the maximum concentration of antibodies in the bloodstream of the patient after the vaccination.
   A second dose of the vaccine has to be given to try to ensure that it is fully effective. It is only safe to give the second dose if the concentration of antibodies in the bloodstream of the patient is less than \(5 \mu \mathrm {~g} / \mathrm { ml }\).
2. Determine whether, according to the model, it is safe to give the second dose of the vaccine to the patient exactly 10 weeks after the first dose.

1. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation

$$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 4 t + 12$$ where \(P\) is \(x\) metres from the origin \(O\) at time \(t\) seconds, \(t \geqslant 0\)

1. Determine the general solution of the differential equation.
2. Hence determine the particular solution for which \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2\) when \(t = 0\)
3. 1. Show that, according to the model, the minimum distance between \(O\) and \(P\) is \(( 2 + \ln 2 )\) metres.
   2. Justify that this distance is a minimum. For large values of \(t\) the particle is expected to move with constant speed.
4. Comment on the suitability of the model in light of this information.