4.10d Second order homogeneous: auxiliary equation method

7. (a) Given that \(x = e ^ { t }\), show that

1. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \frac { \mathrm {~d} y } { \mathrm {~d} t }$$
2. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Use you answers to part (a) to show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = \mathrm { e } ^ { 3 t }$$ (c) Hence find the general solution of $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$

7. (a) Show that the transformation \(x = t ^ { 2 }\) transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 15 y = 15 x$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 15 y = 15 t ^ { 2 }$$ (b) Solve differential equation (II) to determine \(y\) in terms of \(t\).
(c) Hence determine the general solution of differential equation (I). \includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-24_2258_53_308_1980}

6 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y = 2 x + 1$$ Find

1. the complementary function,
2. the general solution. In a particular case, it is given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).
3. Find the solution of the differential equation in this case.
4. Write down the function to which \(y\) approximates when \(x\) is large and positive.

3 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 8 y = \mathrm { e } ^ { 3 x } .$$

8

1. Find the value of the constant \(k\) such that \(y = k x ^ { 2 } \mathrm { e } ^ { - 2 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \mathrm { e } ^ { - 2 x }$$
2. Find the solution of this differential equation for which \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).
3. Use the differential equation to determine the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = 0\). Hence prove that \(0 < y \leqslant 1\) for \(x \geqslant 0\).

1. (a) Show that the transformation \(x = \mathrm { e } ^ { t }\) transforms the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { 2 t }$$ (b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(t\).
(c) Hence find the general solution of the differential equation (I).

6 The differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = \sin k x\) is to be solved, where \(k\) is a constant.

1. In the case \(k = 2\), by using a particular integral of the form \(a x \cos 2 x + b x \sin 2 x\), find the general solution.
2. Describe briefly the behaviour of \(y\) when \(x \rightarrow \infty\).
3. In the case \(k \neq 2\), explain whether \(y\) would exhibit the same behaviour as in part (ii) when \(x \rightarrow \infty\).

6 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 12 \mathrm { e } ^ { 2 x }$$

1. Find the general solution of the differential equation.
2. It is given that the curve which represents a particular solution of the differential equation has gradient 6 when \(x = 0\), and approximates to \(y = \mathrm { e } ^ { 2 x }\) when \(x\) is large and positive. Find the equation of the curve.

5 Find the solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = \mathrm { e } ^ { - x }\) for which \(y = \frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).

5 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = \mathrm { e } ^ { - x }$$ subject to the conditions \(y = \frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).

11 Show that, with a suitable value of the constant \(\alpha\), the substitution \(y = x ^ { \alpha } w\) reduces the differential equation $$2 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( 3 x ^ { 2 } + 8 x \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( x ^ { 2 } + 6 x + 4 \right) y = \mathrm { f } ( x )$$ to $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} w } { \mathrm {~d} x } + w = \mathrm { f } ( x )$$ Find the general solution for \(y\) in the case where \(\mathrm { f } ( x ) = 6 \sin 2 x + 7 \cos 2 x\).

Show that the substitution \(y = x z\) reduces the differential equation $$\frac { 1 } { x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { 6 } { x } - \frac { 2 } { x ^ { 2 } } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + \left( \frac { 9 } { x } - \frac { 6 } { x ^ { 2 } } + \frac { 2 } { x ^ { 3 } } \right) y = 169 \sin 2 x$$ to the differential equation $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 9 z = 169 \sin 2 x$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , z = - 10\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = 5\).

9 Find \(x\) in terms of \(t\) given that $$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 6 \mathrm { e } ^ { - 2 t }$$ and that, when \(t = 0 , x = \frac { 5 } { 3 }\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 7 } { 6 }\). State \(\lim _ { t \rightarrow \infty } x\).

9 Find the value of the constant \(k\) such that \(y = k x ^ { 2 } \mathrm { e } ^ { 2 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 4 \mathrm { e } ^ { 2 x }$$ Hence find the general solution of ( \(*\) ). Find the particular solution of ( \(*\) ) such that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2\) when \(x = 0\).

10 It is given that \(x = t ^ { \frac { 1 } { 2 } }\), where \(x > 0\) and \(t > 0\), and \(y\) is a function of \(x\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x + \frac { 1 } { x } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 2 } y = 4 x ^ { 2 } \mathrm { e } ^ { - x ^ { 2 } }$$ reduces to the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = \mathrm { e } ^ { - t }$$
3. Find the general solution of ( \(*\) ), giving \(y\) in terms of \(x\).

10 It is given that \(t \neq 0\) and $$t \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 t x = 3 t ^ { 2 } + 1$$

1. Show that if \(y = t x\) then $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 9 y = 3 t ^ { 2 } + 1$$
2. Find \(x\) in terms of \(t\), given that \(x = \frac { 1 } { 9 } \pi\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 }\) when \(t = \frac { 1 } { 3 } \pi\).

The variable \(y\) depends on \(x\), and the variables \(x\) and \(t\) are related by \(x = \mathrm { e } ^ { t }\). Show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t } \quad \text { and } \quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } .$$

1. Given that \(y\) satisfies the differential equation $$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 16 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 25 y = 50 ( \ln x ) - 1$$ find a differential equation involving only \(t\) and \(y\).
2. Show that the complementary function of the differential equation in \(t\) and \(y\) may be written in the form $$R \mathrm { e } ^ { - \frac { 3 } { 2 } t } \sin ( 2 t + \phi )$$ where \(R\) and \(\phi\) are arbitrary constants.
3. Find a particular integral of the differential equation in \(t\) and \(y\).
4. Hence find the general solution of the differential equation in \(x\) and \(y\).

9 Show that if \(y\) depends on \(x\) and \(x = \mathrm { e } ^ { u }\) then $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} u } .$$ Given that \(y\) satisfies the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 30 x ^ { 2 }$$ use the substitution \(x = \mathrm { e } ^ { u }\) to show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 3 y = 30 \mathrm { e } ^ { 2 u }$$ Hence find the general solution for \(y\) in terms of \(x\).

11 It is given that \(x \neq 0\) and $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 x y = 8 x ^ { 2 } + 16$$ Show that if \(z = x y\) then $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 z = 8 x ^ { 2 } + 16$$ Find \(y\) in terms of \(x\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2\) when \(x = \frac { 1 } { 2 } \pi\).

6 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = \sin 2 t$$ Describe the behaviour of \(x\) as \(t \rightarrow \infty\), justifying your answer.

3 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 26 t ^ { 2 } + 3 t + 13$$

9 Given that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + ( 2 - 3 x ) y = 10 \mathrm { e } ^ { 2 x }$$ and that \(v = x y\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 3 v = 10 \mathrm { e } ^ { 2 x }$$ Find the general solution for \(y\) in terms of \(x\).

6 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 7 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 10 x = 116 \sin 2 t$$ State an approximate solution for large positive values of \(t\).

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