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Found in: Page 450

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# To derive the general solutions given by equations (17)- (20) for the non-homogeneous equation (16), complete the following steps.(a) Substitute ${\mathbf{y}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{{\mathbf{a}}}_{{\mathbf{n}}}{{\mathbf{x}}}^{{\mathbf{n}}}$ and the Maclaurin series into equation (16) to obtain $\left(2{a}_{2}-{a}_{0}\right){\mathbf{+}}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left[\left(k+2\right)\left(k+1\right){a}_{k+2}-\left(k+1\right){a}_{k}\right]{{\mathbf{x}}}^{{\mathbf{k}}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{\left(}\mathbf{-}\mathbf{1}\mathbf{\right)}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{\right)}\mathbf{!}}{{\mathbf{x}}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$(b) Equate the coefficients of like powers on both sides of the equation in part (a) and thereby deduce the equations${{\mathbf{a}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}{\mathbf{,}}{{\mathbf{a}}}_{{\mathbf{3}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{6}}{\mathbf{+}}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{3}}{\mathbf{,}}{{\mathbf{a}}}_{{\mathbf{4}}}{\mathbf{=}}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{8}}{\mathbf{,}}{{\mathbf{a}}}_{{\mathbf{5}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{40}}{\mathbf{+}}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{15}}{\mathbf{,}}{{\mathbf{a}}}_{{\mathbf{6}}}{\mathbf{=}}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{48}}{\mathbf{,}}{{\mathbf{a}}}_{{\mathbf{7}}}{\mathbf{=}}\frac{\mathbf{19}}{\mathbf{5040}}{\mathbf{+}}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{105}}$ (c) Show that the relations in part (b) yield the general solution to (16) given in equations (17)-(20).

(a) After substituting $\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ and Maclaurin series, we get the solution $\left({\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\sum _{\mathrm{k}=1}^{\infty }\left[\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right){\mathrm{a}}_{\mathrm{k}+2}-\left(\mathrm{k}+1\right){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots$.

(b) After equating the coefficients, the deduced equations are

$2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}=0\to {\mathrm{a}}_{2}=\frac{{\mathrm{a}}_{0}}{2}\phantom{\rule{0ex}{0ex}}6{\mathrm{a}}_{3}-2{\mathrm{a}}_{1}=1\to {\mathrm{a}}_{3}=\frac{1}{6}+\frac{{\mathrm{a}}_{1}}{3}\phantom{\rule{0ex}{0ex}}12{\mathrm{a}}_{4}-3{\mathrm{a}}_{2}=0\to {\mathrm{a}}_{4}=\frac{{\mathrm{a}}_{0}}{8}\phantom{\rule{0ex}{0ex}}20{\mathrm{a}}_{5}-4{\mathrm{a}}_{3}=-\frac{1}{6}\to {\mathrm{a}}_{5}=\frac{1}{40}+\frac{{\mathrm{a}}_{1}}{15}\phantom{\rule{0ex}{0ex}}30{\mathrm{a}}_{6}-5{\mathrm{a}}_{4}=0\to {\mathrm{a}}_{6}=\frac{{\mathrm{a}}_{0}}{48}\phantom{\rule{0ex}{0ex}}42{\mathrm{a}}_{7}-6{\mathrm{a}}_{5}=\frac{1}{120}\to {\mathrm{a}}_{7}=\frac{19}{5040}+\frac{{\mathrm{a}}_{1}}{105}$

(c) We showed that the results are similar to equations (17)- (20).

See the step by step solution

## Step 1: Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

${\mathbf{y}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{{\mathbf{a}}}_{{\mathbf{n}}}{{\mathbf{x}}}^{{\mathbf{n}}}$

## Step 2: Find the relation.

The equation given in (16) is:

$\mathrm{y}\text{'}\text{'}-\mathrm{xy}\text{'}-\mathrm{y}=\mathrm{sinx}$

Use the formula

$\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$

Take derivative

$\mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\phantom{\rule{0ex}{0ex}}\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}$

The Maclaurin series is $\mathrm{sinx}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots$

Substitute in the above equation.

role="math" localid="1664105464082" $\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}-\mathrm{x}·\sum _{\mathrm{n}=1}^{\infty }{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots \phantom{\rule{0ex}{0ex}}\sum _{\mathrm{k}=0}^{\infty }\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{ka}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=0}^{\infty }{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots \phantom{\rule{0ex}{0ex}}{\mathrm{a}}_{2}+\sum _{\mathrm{k}=1}^{\infty }\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{ka}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-{\mathrm{a}}_{0}-\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots \phantom{\rule{0ex}{0ex}}\left({\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\sum _{\mathrm{k}=1}^{\infty }\left[\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right){\mathrm{a}}_{\mathrm{k}+2}-\left(\mathrm{k}+1\right){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots$

Hence the relation is $\left({\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\sum _{\mathrm{k}=1}^{\infty }\left[\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right){\mathrm{a}}_{\mathrm{k}+2}-\left(\mathrm{k}+1\right){\mathrm{a}}_{\mathrm{k}}\right]{\mathrm{x}}^{\mathrm{k}}=\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{5}}{5!}-\cdots$.

## Step 3: Find the equations of coefficients.

Expand the expression.

$\left(2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}-2{\mathrm{a}}_{1}\right)\mathrm{x}+\left(12{\mathrm{a}}_{4}-3{\mathrm{a}}_{3}\right){\mathrm{x}}^{2}++\left(20{\mathrm{a}}_{5}-4{\mathrm{a}}_{3}\right){\mathrm{x}}^{3}+\left(30{\mathrm{a}}_{6}-5{\mathrm{a}}_{4}\right){\mathrm{x}}^{4}\phantom{\rule{0ex}{0ex}}+\left(42{\mathrm{a}}_{7}-6{\mathrm{a}}_{5}\right){\mathrm{x}}^{5}+\left(56{\mathrm{a}}_{8}-6{\mathrm{a}}_{6}\right){\mathrm{x}}^{6}+\cdots =\mathrm{x}-\frac{{\mathrm{x}}^{3}}{6}+\frac{{\mathrm{x}}^{5}}{120}-\frac{{\mathrm{x}}^{7}}{5040}+\cdots$

Equate the coefficients of like powers on both sides of the equation.

$2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}=0\to {\mathrm{a}}_{2}=\frac{{\mathrm{a}}_{0}}{2}\phantom{\rule{0ex}{0ex}}6{\mathrm{a}}_{3}-2{\mathrm{a}}_{1}=1\to {\mathrm{a}}_{3}=\frac{1}{6}+\frac{{\mathrm{a}}_{1}}{3}\phantom{\rule{0ex}{0ex}}12{\mathrm{a}}_{4}-3{\mathrm{a}}_{2}=0\to {\mathrm{a}}_{4}=\frac{{\mathrm{a}}_{0}}{8}\phantom{\rule{0ex}{0ex}}20{\mathrm{a}}_{5}-4{\mathrm{a}}_{3}=-\frac{1}{6}\to {\mathrm{a}}_{5}=\frac{1}{40}+\frac{{\mathrm{a}}_{1}}{15}\phantom{\rule{0ex}{0ex}}30{\mathrm{a}}_{6}-5{\mathrm{a}}_{4}=0\to {\mathrm{a}}_{6}=\frac{{\mathrm{a}}_{0}}{48}\phantom{\rule{0ex}{0ex}}42{\mathrm{a}}_{7}-6{\mathrm{a}}_{5}=\frac{1}{120}\to {\mathrm{a}}_{7}=\frac{19}{5040}+\frac{{\mathrm{a}}_{1}}{105}$

## Step 4: Find the expression after expansion.

Substitute the above coefficients in the equation:

$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{x}+\frac{{\mathrm{a}}_{0}}{2}{\mathrm{x}}^{2}+\left(\frac{1}{6}+\frac{{\mathrm{a}}_{1}}{3}\right){\mathrm{x}}^{3}+\frac{{\mathrm{a}}_{0}}{8}{\mathrm{x}}^{4}+\left(\frac{1}{40}+\frac{{\mathrm{a}}_{1}}{15}\right){\mathrm{x}}^{5}+\frac{{\mathrm{a}}_{0}}{48}{\mathrm{x}}^{6}+\left(\frac{19}{5040}+\frac{{\mathrm{a}}_{1}}{105}\right){\mathrm{x}}^{7}+\cdots \phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_{0}\left(1+\frac{{\mathrm{x}}^{2}}{2}+\frac{{\mathrm{x}}^{4}}{8}+\frac{{\mathrm{x}}^{6}}{48}+\cdots \right)+{\mathrm{a}}_{1}\left(\mathrm{x}+\frac{{\mathrm{x}}^{3}}{3}+\frac{{\mathrm{x}}^{5}}{15}+\frac{{\mathrm{x}}^{7}}{105}+\cdots \right)+\frac{{\mathrm{x}}^{3}}{6}+\frac{{\mathrm{x}}^{5}}{40}+\frac{19{\mathrm{x}}^{7}}{5040}+\cdots \phantom{\rule{0ex}{0ex}}$

Where we can write:

${\mathrm{y}}_{1}=1+\frac{{\mathrm{x}}^{2}}{2}+\frac{{\mathrm{x}}^{4}}{8}+\frac{{\mathrm{x}}^{6}}{48}+\cdots \phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{2}=\mathrm{x}+\frac{{\mathrm{x}}^{3}}{3}+\frac{{\mathrm{x}}^{5}}{15}+\frac{{\mathrm{x}}^{7}}{105}+\cdots \phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{p}}=\frac{{\mathrm{x}}^{3}}{6}+\frac{{\mathrm{x}}^{5}}{40}+\frac{19{\mathrm{x}}^{7}}{5040}+\cdots$

Hence, we showed that the results are similar to equations (17)-(20).