4.10e Second order non-homogeneous: complementary + particular integral

2

1. Find the values of the constants \(p\) and \(q\) for which \(p + q x \mathrm { e } ^ { - 2 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = 4 - 9 \mathrm { e } ^ { - 2 x }$$
2. Hence find the general solution of this differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 4\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0\) as \(x \rightarrow \infty\).

7

1. Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$\text { into } \quad \begin{aligned} x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x ) \\ \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t \end{aligned}$$ (7 marks)
2. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$ (11 marks)
3. Write down the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$

3

1. Find the values of the constants \(a , b\) and \(c\) for which \(a + b x + c x \mathrm { e } ^ { - 3 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 3 x - 8 \mathrm { e } ^ { - 3 x }$$
2. Hence find the general solution of this differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 1\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow - 1\) as \(x \rightarrow \infty\).

7 A differential equation is given by $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x , \quad 0 < x < \pi$$

1. Show that the substitution $$y = u \sin x$$ where \(u\) is a function of \(x\), transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + u = \sin 2 x$$
2. Hence find the general solution of the differential equation $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (6 marks)

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}

5

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$
2. It is given that \(y = \mathrm { f } ( x )\) is the solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$ such that \(\mathrm { f } ( 0 ) = 0\) and \(\mathrm { f } ^ { \prime } ( 0 ) = 0\).
   1. Show that \(f ^ { \prime \prime } ( 0 ) = 0\).
   2. Find the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\mathrm { f } ( x )\).
      [0pt] [3 marks]

6 A differential equation is given by $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$

1. Show that the substitution \(x = \mathrm { e } ^ { 2 t }\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = 4 t ^ { 2 } + 5 \mathrm { e } ^ { - t }$$
2. Hence find the general solution of the differential equation $$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$
   \includegraphics[max width=\textwidth, alt={}]{7b4a1237-bb28-4cba-84b1-35fa91d87408-14_1634_1709_1071_153}

1

1. Find the values of the constants \(a\) and \(b\) for which \(a x + b\) is a particular integral of the differential equation $$2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 5 y = 10 x$$
2. Hence find the general solution of \(2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 5 y = 10 x\).
   [0pt] [3 marks]

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

1. Find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = 0\).
2. Hence find the general solution of the differential equation \(\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 5 y = x ( 4 - 5 x )\).

11 A company is required to weigh any goods before exporting them overseas. When a crate is placed on a set of weighing scales, the mass displayed takes time to settle down to its final value. The company wishes to model the mass, \(m \mathrm {~kg}\), which is displayed \(t\) seconds after a crate X is placed on the scales.
For the displayed mass it is assumed that the rate of change of the quantity \(\left( 0.5 \frac { \mathrm {~d} m } { \mathrm {~d} t } + m \right)\) with respect to time is proportional to \(( 80 - m )\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 2 \mathrm {~km} = 160 \mathrm { k }\), where \(k\) is a real constant. It is given that the complementary function for the differential equation in part (i) is \(\mathrm { e } ^ { \lambda t } ( A \cos 2 t + B \sin 2 t )\), where \(A\) and \(B\) are arbitrary constants.
2. Show that \(k = \frac { 5 } { 2 }\) and state the value of the constant \(\lambda\). When X is initially placed on the scales the displayed mass is zero and the rate of increase of the displayed mass is \(160 \mathrm {~kg} \mathrm {~s} ^ { - 1 }\).
3. Find \(m\) in terms of \(t\).
4. Describe the long term behaviour of \(m\).
5. With reference to your answer to part (iv), comment on a limitation of the model.
6. (a) Find the value of \(m\) that corresponds to the stationary point on the curve \(m = \mathrm { f } ( t )\) with the smallest positive value of \(t\).
   (b) Interpret this value of \(m\) in the context of the model.
7. Adapt the differential equation \(\frac { \mathrm { d } ^ { 2 } m } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} m } { \mathrm {~d} t } + 5 m = 400\) to model the mass displayed \(t\) seconds after a crate Y , of mass 100 kg , is placed on the scales. \section*{END OF QUESTION PAPER} \section*{Copyright Information:} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
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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.

6. \begin{figure}[h] \end{figure} A particle \(P\) of mass 2 kg is attached to the mid-point of a light elastic spring of natural length 2 m and modulus of elasticity 4 N . One end \(A\) of the elastic spring is attached to a fixed point on a smooth horizontal table. The spring is then stretched until its length is 4 m and its other end \(B\) is held at a point on the table where \(A B = 4 \mathrm {~m}\). At time \(t = 0 , P\) is at rest on the table at the point \(O\) where \(A O = 2 \mathrm {~m}\), as shown in Fig. 3. The end \(B\) is now moved on the table in such a way that \(A O B\) remains a straight line. At time \(t\) seconds, \(A B = \left( 4 + \frac { 1 } { 2 } \sin 4 t \right) \mathrm { m }\) and \(A P = ( 2 + x ) \mathrm { m }\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 x = \sin 4 t$$
2. Hence find the time when \(P\) first comes to instantaneous rest. END

5. A light elastic string, of natural length \(2 a\) and modulus of elasticity \(m g\), has a particle \(P\) of mass \(m\) attached to its mid-point. One end of the string is attached to a fixed point \(A\) and the other end is attached to a fixed point \(B\) which is at a distance \(4 a\) vertically below \(A\).

1. Show that \(P\) hangs in equilibrium at the point \(E\) where \(A E = \frac { 5 } { 2 } a\). The particle \(P\) is held at a distance \(3 a\) vertically below \(A\) and is released from rest at time \(t = 0\). When the speed of the particle is \(v\), there is a resistance to motion of magnitude \(2 m k v\), where \(k = \sqrt { } \left( \frac { g } { a } \right)\). At time \(t\) the particle is at a distance \(\left( \frac { 5 } { 2 } a + x \right)\) from \(A\).
2. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = 0$$
3. Hence find \(x\) in terms of \(t\).

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 \(2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)\).

5. A light elastic spring has natural length \(l\) and modulus of elasticity \(m g\). One end of the spring is fixed to a point \(O\) on a rough horizontal table. The other end is attached to a particle \(P\) of mass \(m\) which is at rest on the table with \(O P = l\). At time \(t = 0\) the particle is projected with speed \(\sqrt { } ( g l )\) along the table in the direction \(O P\). At time \(t\) the displacement of \(P\) from its initial position is \(x\) and its speed is \(v\). The motion of \(P\) is subject to air resistance of magnitude \(2 m v \omega\), where \(\omega = \sqrt { \frac { g } { l } }\). The coefficient of friction between \(P\) and the table is 0.5 .

1. Show that, until \(P\) first comes to rest, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + \omega ^ { 2 } x = - 0.5 g$$
2. Find \(x\) in terms of \(t , l\) and \(\omega\).
3. Hence find, in terms of \(\omega\), the time taken for \(P\) to first come to instantaneous rest.
   (3)

6. A light elastic spring \(A B\) has natural length \(2 a\) and modulus of elasticity \(2 m n ^ { 2 } a\), where \(n\) is a constant. A particle \(P\) of mass \(m\) is attached to the end \(A\) of the spring. At time \(t = 0\), the spring, with \(P\) attached, lies at rest and unstretched on a smooth horizontal plane. The other end \(B\) of the spring is then pulled along the plane in the direction \(A B\) with constant acceleration \(f\). At time \(t\) the extension of the spring is \(x\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
2. Find \(x\) in terms of \(n , f\) and \(t\). Hence find
3. the maximum extension of the spring,
4. the speed of \(P\) when the spring first reaches its maximum extension.
   \section*{June 2009}

A particle of mass \(m\) is projected vertically upwards, at time \(t = 0\), with speed \(U\). The particle is subject to air resistance of magnitude \(\frac { m g v ^ { 2 } } { k ^ { 2 } }\), where \(v\) is the speed of the particle at time \(t\) and \(k\) is a positive constant.

1. Show that the particle reaches its greatest height above the point of projection at time $$\frac { k } { g } \tan ^ { - 1 } \left( \frac { U } { k } \right)$$
2. Find the greatest height above the point of projection attained by the particle.

A particle \(P\) of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point \(O\) on the line. At time \(t\) seconds, \(P\) is \(x\) metres from \(O\) and the force towards \(O\) has magnitude \(9 x\) newtons. The particle \(P\) is also subject to air resistance, which has magnitude \(12 v\) newtons when \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that the equation of motion of \(P\) is $$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 12 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 0$$ It is given that the solution of this differential equation is of the form $$x = \mathrm { e } ^ { - \lambda t } ( A t + B )$$ When \(t = 0\) the particle is released from rest at the point \(R\), where \(O R = 4 \mathrm {~m}\). Find,
2. the values of the constants \(\lambda , A\) and \(B\),
3. the greatest speed of \(P\) in the subsequent motion.
   

7. A particle \(P\) of mass 0.5 kg is attached to the end \(A\) of a light elastic spring \(A B\), of natural length 0.6 m and modulus of elasticity 2.7 N . At time \(t = 0\) the end \(B\) of the spring is held at rest and \(P\) hangs at rest at the point \(C\) which is vertically below \(B\). The end \(B\) is then moved along the line of the spring so that, at time \(t\) seconds, the downwards displacement of \(B\) from its initial position is \(4 \sin 2 t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of \(P\) below \(C\) is \(y\) metres.

1. Show that $$y + \frac { 49 } { 45 } = x + 4 \sin 2 t$$
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 36 \sin 2 t$$ Given that \(y = \frac { 36 } { 5 } \sin 2 t\) is a particular integral of this differential equation,
3. find \(y\) in terms of \(t\),
4. find the speed of \(P\) when \(t = \frac { 1 } { 3 } \pi\).

5. \begin{figure}[h] \end{figure} A particle \(P\) of mass 1.5 kg is attached to the midpoint of a light elastic spring \(A B\), of natural length 2 m and modulus of elasticity 12 N . The end \(A\) of the spring is attached to a fixed point on a smooth horizontal floor. The end \(B\) is held at a point on the floor where \(A B = 6 \mathrm {~m}\). At time \(t = 0 , P\) is at rest on the floor at the point \(O\), where \(A O = 3 \mathrm {~m}\), as shown in Figure 2. The end \(B\) is now moved along the floor in such a way that \(A O B\) remains a straight line and at time \(t\) seconds, \(t \geqslant 0\), $$A B = \left( 6 + \frac { 1 } { 4 } \sin 2 t \right) \mathrm { m }$$ At time \(t\) seconds, \(A P = ( 3 + x ) \mathrm { m }\).

1. Show that, for \(t \geqslant 0\), $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 16 x = 2 \sin 2 t$$ The general solution of this differential equation is $$x = C \cos 4 t + D \sin 4 t + \frac { 1 } { 6 } \sin 2 t$$ where \(C\) and \(D\) are constants.
2. Find the time at which \(P\) first comes to instantaneous rest. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{44066c44-e366-4f87-b1b2-c5a894e407fa-20_705_1104_116_420} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure}

4. A particle \(P\) of mass 9 kg moves along the horizontal positive \(x\)-axis under the action of a force directed towards the origin. At time \(t\) seconds, the displacement of \(P\) from \(O\) is \(x\) metres, \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the force has magnitude \(16 x\) newtons. The particle \(P\) is also subject to a resistive force of magnitude \(24 v\) newtons.

1. Show that the equation of motion of \(P\) is $$9 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 24 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 16 x = 0$$ It is given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - \frac { 4 } { 3 } t } ( A t + B )$$ where \(A\) and \(B\) are arbitrary constants.
   When \(t = \frac { 3 } { 4 } , P\) is travelling towards \(O\) with its maximum speed of \(8 \mathrm { e } ^ { - 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(x = d\).
2. Find the value of \(d\).
3. Find the value of \(x\) when \(t = 0\)

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 differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$ When \(t = 0\), the velocity of \(P\) is \(( 8 \mathbf { i } - 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\).
Find the velocity of \(P\) when \(t = \frac { 2 } { 3 } \ln 2\).
(7)