Q32P Using Problems 29 聽and 31b show... [FREE SOLUTION]

Chapter 8: Q32P (page 425)

Using Problems 29 and 31b show that equation (6.24) is correct.

Short Answer

Expert verified

Answer

It is proved that $y_{p} = \{\begin{cases} e^{c x} Q_{n} (x) i f c 鈮� t o e i t h e r a o r b \\ x e^{c x} Q_{n} (x) i f c = t o e i t h e r a o r b, a 鈮� b \\ x^{2} e^{c x} Q_{n} (x) i f c = a = b \end{cases}$

Step by step solution

Given data.

Information given in the question is $(D - a) (D - b) y = \{\begin{matrix} k \sin a x \\ k \cos a x \end{matrix}\}$

Condition to prove the statement.

Here $y_{p}$ is a polynomial $Q_{n} (x)$ of degree $n$

$y_{p} = {\begin{cases} e^{c x} Q_{n} (x) i f c 鈮� t o e i t h e r a o r b \\ x e^{c x} Q_{n} (x) i f c = t o e i t h e r a o r b, a 鈮� b \\ x^{2} e^{c x} Q_{n} (x) i f c = a = b \end{cases}$

Find the particular solution of (D-a)(D-b)y={k sin axk cos ax}

A particular solution of $(D - a) (D - b) y = \{\begin{matrix} k \sin a x \\ k \cos a x \end{matrix}\}$

First solve

$(D - a) (D - b) y = k e^{k i x} (D - a) (D - b) y = k (c o s a x + i n i n a x) (D - a) (D - b) y = (k c o s a x + K i n i n a x)$

Take the real and imaginary part

$(D - a) (D - b) y = \{\begin{matrix} k \sin a x \\ k \cos a x \end{matrix}\}$

A particular solution role="math" localid="1654061711166" $y_{P}$ of $(D - a) (D - b) y = e^{c x} P_{n} (x)$

Where $P_{n} (x)$ is a polynomial of degree n is

$y_{p} = {\begin{cases} e^{c x} Q_{n} (x) i f c 鈮� t o e i t h e r a o r b \\ x e^{c x} Q_{n} (x) i f c = t o e i t h e r a o r b, a 鈮� b \\ x^{2} e^{c x} Q_{n} (x) i f c = a = b \end{cases}$

Here $Q_{n} (x)$ is a polynomial of the degree as $P_{n} (x)$ with undetermined coefficients to be found to satisfy the given differential equation.

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91影视

Using Problems 29 and 31b show that equation (6.24) is correct.

Short Answer

Step by step solution

Given data.

Condition to prove the statement.

Find the particular solution of (D-a)(D-b)y={k sin axk cos ax}

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