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

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 $n$

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

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

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

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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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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