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

6. (a) Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 3 x ^ { 2 } + 2 x + 1$$ (9)
(b) Find the particular solution of this differential equation for which \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\) (5)

4. $$y = 3 \mathrm { e } ^ { - x } \cos 3 x + A \mathrm { e } ^ { - x } \sin 3 x$$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 40 \mathrm { e } ^ { - x } \sin 3 x$$ where \(A\) is a constant.

1. Find the value of \(A\).
2. Hence find the general solution of this differential equation.
3. Find the particular solution of this differential equation for which both \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) at \(x = 0\)

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.

1. (a) Given that \(t = \ln x\), where \(x > 0\), show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Hence show that the transformation \(t = \ln x\), where \(x > 0\), transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 y = 1 + 4 \ln x - 2 ( \ln x ) ^ { 2 }$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } - 2 y = 1 + 4 t - 2 t ^ { 2 }$$ (c) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(d) Hence determine the general solution of differential equation (I).

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t ^ { 2 } + 11 t$$ (b) Find the particular solution of this differential equation for which \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = 1\) when \(t = 0\).
(c) For this particular solution, calculate the value of \(y\) when \(t = 1\).

8. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 9 y = 4 \mathrm { e } ^ { 3 t } , \quad t \geq 0 .$$

1. Show that \(K t ^ { 2 } e ^ { 3 t }\) is a particular integral of the differential equation, where \(K\) is a constant to be found.
2. Find the general solution of the differential equation. (3) Given that a particular solution satisfies \(\boldsymbol { y } = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = 1\) when \(\boldsymbol { t } = \mathbf { 0 }\),
3. find this solution.(4) Another particular solution which satisfies \(\boldsymbol { y } = \mathbf { 1 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = \mathbf { 0 }\) when \(\boldsymbol { t } = \mathbf { 0 }\), has equation $$y = \left( 1 - 3 t + 2 t ^ { 2 } \right) \mathrm { e } ^ { 3 t }$$
4. For this particular solution draw a sketch graph of \(y\) against \(t\), showing where the graph crosses the \(t\)-axis. Determine also the coordinates of the minimum of the point on the sketch graph.

4. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 65 \sin 2 x , x > 0$$

1. Find the general solution of the differential equation.
2. Show that for large values of \(x\) this general solution may be approximated by a sine function and find this sine function.
   (3)(Total 12 marks)

8. (a) Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = 2 \mathrm { e } ^ { - t }$$ (b) Find the particular solution that satisfies \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = 1\) at \(t = 0\).
(6)(Total 12 marks)

3. (a) Show that the transformation \(y = x v\) transforms the equation $$\begin{array} { l l } x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + \left( 2 + 9 x ^ { 2 } \right) y = x ^ { 5 } , \\ \text { into the equation } & \frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 9 v = x ^ { 2 } . \end{array}$$I (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.
(1)(Total 12 marks)

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 2 t + 9$$ (b) Find the particular solution of this differential equation for which \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 1\) when \(t = 0\). The particular solution in part (b) is used to model the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds \(( t \geq 0 ) , P\) is \(x\) metres from the origin \(O\).
(c) Show that the minimum distance between \(O\) and \(P\) is \(\frac { 1 } { 2 } ( 5 + \ln 2 ) \mathrm { m }\) and justify that the distance is a minimum.
(4)(Total 14 marks)

1. Given that \(3 x \sin 2 x\) is a particular integral of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = k \cos 2 x$$ where \(k\) is a constant,

1. calculate the value of \(k\),
2. find the particular solution of the differential equation for which at \(x = 0 , y = 2\), and for which at \(x = \frac { \pi } { 4 } , y = \frac { \pi } { 2 }\).
   (4)(Total 8 marks)

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 \(0 \leqslant x \leqslant \pi\).

1. The differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = \cos 3 t , \quad t \geqslant 0$$ describes the motion of a particle along the \(x\)-axis.

1. Find the general solution of this differential equation.
2. Find the particular solution of this differential equation for which, at \(t = 0\), $$x = \frac { 1 } { 2 } \text { and } \frac { \mathrm { d } x } { \mathrm {~d} t } = 0$$ On the graph of the particular solution defined in part (b), the first turning point for \(t > 30\) is the point \(A\).
3. Find approximate values for the coordinates of \(A\).

4. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 6 x = 2 \cos t - \sin t$$

1. (a) Find the value of \(\lambda\) for which \(\lambda t ^ { 2 } \mathrm { e } ^ { 3 t }\) is a particular integral of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 9 y = 6 \mathrm { e } ^ { 3 t } , \quad t \geqslant 0$$ (b) Hence find the general solution of this differential equation. Given that when \(t = 0 , y = 5\) and \(\frac { \mathrm { d } y } { \mathrm {~d} t } = 4\) (c) find the particular solution of this differential equation, giving your solution in the form \(y = \mathrm { f } ( t )\).

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 $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 y = 6 \mathrm { e } ^ { 2 x }$$

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 \(y = \mathrm { f } ( x )\). \(\square\)

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

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 27 \mathrm { e } ^ { - x }$$ (b) Find the particular solution that satisfies \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\)