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

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

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 7 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 16 y = 2 \ln x , \quad x > 0$$ into the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 8 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 16 y = 2 u$$ (b) Find the general solution of the differential equation (II), expressing \(y\) as a function of \(u\).
(c) Hence obtain the general solution of the differential equation (I).

7. (a) Show that the substitution \(x = e ^ { u }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = - x ^ { - 2 } , \quad x > 0$$ into the equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 2 y = - \mathrm { e } ^ { - 2 u }$$ (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 \(y = \mathrm { f } ( x )\)

5. (a) Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 26 \sin 3 x$$ (b) Find the particular solution of this differential equation for which \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\) when \(x = 0\)

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

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}

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

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 2 \sin x$$ Given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = 0\) (b) find the particular solution of differential equation (I).

1. (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 \(0 \leqslant x \leqslant \pi\).

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).

4 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 65 \sin 2 x$$

5 The variables \(x\) and \(y\) satisfy the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 y = 5 \mathrm { e } ^ { - 2 x }$$

1. Find the complementary function of the differential equation.
2. Given that there is a particular integral of the form \(y = p x \mathrm { e } ^ { - 2 x }\), find the constant \(p\).
3. Find the solution of the equation for which \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4\) when \(x = 0\).

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 \(x = 0\).

1 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 13 y = \sin x$$

5 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = \mathrm { e } ^ { 3 x }$$

1. Find the complementary function.
2. Explain briefly why there is no particular integral of either of the forms \(y = k \mathrm { e } ^ { 3 x }\) or \(y = k x \mathrm { e } ^ { 3 x }\).
3. Given that there is a particular integral of the form \(y = k x ^ { 2 } \mathrm { e } ^ { 3 x }\), find the value of \(k\).

5 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 85 \cos x .$$

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 \(\mathrm { f } ( x ) = 6 \sin 2 x + 7 \cos 2 x\).

8 Find the general solution of the differential equation $$4 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 65 y = 65 x ^ { 2 } + 8 x + 73$$ Show that, whatever the initial conditions, \(\frac { y } { x ^ { 2 } } \rightarrow 1\) as \(x \rightarrow \infty\).

The variables \(z\) and \(x\) are related by the differential equation $$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$ Use the substitution \(y = z ^ { 3 }\) to show that \(y\) and \(x\) are related by the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5 x + 2$$ Given that \(z = 1\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }\) when \(x = 0\), find \(z\) in terms of \(x\). Deduce that, for large positive values of \(x , z \approx x ^ { \frac { 1 } { 3 } }\).

8 Obtain the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 10 \sin 3 x - 20 \cos 3 x$$ Show that, for large positive \(x\) and independently of the initial conditions, $$y \approx R \sin ( 3 x + \phi )$$ where the constants \(R\) and \(\phi\), such that \(R > 0\) and \(0 < \phi < 2 \pi\), are to be determined correct to 2 decimal places.