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Expert-verified Found in: Page 449 ### 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 # In Problems 13-19, find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{ty}}{\mathbf{\text{'}}}{\mathbf{+}}{{\mathbf{e}}}^{{\mathbf{t}}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}{\mathbf{}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}$

The first four nonzero terms in the power series expansion of the given initial value problem $\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}+{\mathrm{e}}^{\mathrm{t}}\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^{3}}{3}+\frac{{\mathrm{t}}^{4}}{12}-\frac{{\mathrm{t}}^{5}}{24}$.

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.

Given,

$\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}+{\mathrm{e}}^{\mathrm{t}}\mathrm{y}=0;\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=-1$

Use the formula,

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

Taking derivative and substituting in the equation, we get the relation,

$\mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}\phantom{\rule{0ex}{0ex}}\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}\phantom{\rule{0ex}{0ex}}\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+\mathrm{t}·\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}+{\mathrm{e}}^{\mathrm{t}}·\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$

Hence we get the relation $\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+\mathrm{t}·\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}+{\mathrm{e}}^{\mathrm{t}}·\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$.

## Step 3: Find the expression after expansion.

The series expansion for the function is

$\left(2{\mathrm{a}}_{2}+6{\mathrm{a}}_{3}\mathrm{t}+12{\mathrm{a}}_{4}{\mathrm{t}}^{2}+20{\mathrm{a}}_{5}{\mathrm{t}}^{3}+30{\mathrm{a}}_{6}{\mathrm{t}}^{4}+\cdots \right)+\mathrm{t}·\left({\mathrm{a}}_{1}+2{\mathrm{a}}_{2}\mathrm{t}+3{\mathrm{a}}_{3}{\mathrm{t}}^{2}+4{\mathrm{a}}_{4}{\mathrm{t}}^{4}+5{\mathrm{a}}_{5}{\mathrm{t}}^{4}+\cdots \right)\phantom{\rule{0ex}{0ex}}+\left(1+\mathrm{t}+\frac{{\mathrm{t}}^{2}}{2}+\frac{{\mathrm{t}}^{3}}{6}+\cdots \right)\left({\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{t}+{\mathrm{a}}_{2}{\mathrm{t}}^{2}+{\mathrm{a}}_{3}{\mathrm{t}}^{3}+\cdots \right)=0$

By expanding the series we get,

role="math" localid="1664095204500" $\left(2{\mathrm{a}}_{2}+6{\mathrm{a}}_{3}\mathrm{t}+12{\mathrm{a}}_{4}{\mathrm{t}}^{2}+20{\mathrm{a}}_{5}{\mathrm{t}}^{3}+30{\mathrm{a}}_{6}{\mathrm{t}}^{4}+\cdots \right)+\left({\mathrm{a}}_{1}\mathrm{t}+2{\mathrm{a}}_{2}{\mathrm{t}}^{2}+3{\mathrm{a}}_{3}{\mathrm{t}}^{3}+4{\mathrm{a}}_{4}{\mathrm{t}}^{5}+5{\mathrm{a}}_{5}{\mathrm{t}}^{6}+\cdots \right)\phantom{\rule{0ex}{0ex}}+\left({\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{t}+{\mathrm{a}}_{2}{\mathrm{t}}^{2}+{\mathrm{a}}_{3}{\mathrm{t}}^{3}+\cdots \right)+\left({\mathrm{a}}_{0}\mathrm{t}+{\mathrm{a}}_{1}{\mathrm{t}}^{2}+{\mathrm{a}}_{2}{\mathrm{t}}^{3}+{\mathrm{a}}_{3}{\mathrm{t}}^{4}+\cdots \right)+\left({\mathrm{a}}_{0}\frac{{\mathrm{t}}^{2}}{2}+{\mathrm{a}}_{1}\frac{{\mathrm{t}}^{3}}{2}+{\mathrm{a}}_{2}\frac{{\mathrm{t}}^{4}}{2}+{\mathrm{a}}_{3}\frac{{\mathrm{t}}^{5}}{2}+\cdots \right)\phantom{\rule{0ex}{0ex}}+\left({\mathrm{a}}_{0}\frac{{\mathrm{t}}^{3}}{6}+{\mathrm{a}}_{1}\frac{{\mathrm{t}}^{4}}{6}+{\mathrm{a}}_{2}\frac{{\mathrm{t}}^{5}}{6}+{\mathrm{a}}_{3}\frac{{\mathrm{t}}^{6}}{6}+\cdots \right)+\cdots =0$

Simplify the expression.

$\left(2{\mathrm{a}}_{2}+{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}+{\mathrm{a}}_{1}+{\mathrm{a}}_{1}+{\mathrm{a}}_{0}\right)\mathrm{t}+\left(12{\mathrm{a}}_{4}+{\mathrm{a}}_{2}+{\mathrm{a}}_{2}+{\mathrm{a}}_{1}+\frac{{\mathrm{a}}_{0}}{2}\right){\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}+\left(20{\mathrm{a}}_{5}+3{\mathrm{a}}_{3}+{\mathrm{a}}_{3}+{\mathrm{a}}_{2}+\frac{{\mathrm{a}}_{1}}{2}+\frac{{\mathrm{a}}_{0}}{6}+\cdots \right){\mathrm{t}}^{3}+\cdots =0$

Hence, the expression after the expansion is:

$\left(2{\mathrm{a}}_{2}+{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}+{\mathrm{a}}_{1}+{\mathrm{a}}_{1}+{\mathrm{a}}_{0}\right)\mathrm{t}+\left(12{\mathrm{a}}_{4}+{\mathrm{a}}_{2}+{\mathrm{a}}_{2}+{\mathrm{a}}_{1}+\frac{{\mathrm{a}}_{0}}{2}\right){\mathrm{t}}^{2}\phantom{\rule{0ex}{0ex}}+\left(20{\mathrm{a}}_{5}+3{\mathrm{a}}_{3}+{\mathrm{a}}_{3}+{\mathrm{a}}_{2}+\frac{{\mathrm{a}}_{1}}{2}+\frac{{\mathrm{a}}_{0}}{6}+\cdots \right){\mathrm{t}}^{3}+\cdots =0$

## Step 4: Find the first four nonzero terms.

By equating the coefficients, we get,

$6{\mathrm{a}}_{3}+{\mathrm{a}}_{1}+{\mathrm{a}}_{1}+{\mathrm{a}}_{0}=0\to {\mathrm{a}}_{3}=-\frac{{\mathrm{a}}_{1}}{3}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}12{\mathrm{a}}_{4}+{\mathrm{a}}_{2}+{\mathrm{a}}_{2}+{\mathrm{a}}_{1}+\frac{{\mathrm{a}}_{0}}{2}=0\to {\mathrm{a}}_{4}=\frac{-{\mathrm{a}}_{0}}{12}=\frac{1}{12}\phantom{\rule{0ex}{0ex}}20{\mathrm{a}}_{5}+3{\mathrm{a}}_{3}+{\mathrm{a}}_{3}+{\mathrm{a}}_{2}+\frac{{\mathrm{a}}_{1}}{2}+\frac{{\mathrm{a}}_{0}}{6}\to {\mathrm{a}}_{5}=\frac{-4{\mathrm{a}}_{3}-\frac{{\mathrm{a}}_{1}}{2}}{20}=-\frac{1}{24}$

The general solution was

$\mathrm{y}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{t}+{\mathrm{a}}_{2}{\mathrm{t}}^{2}+{\mathrm{a}}_{3}{\mathrm{t}}^{3}+\cdots$

Apply the initial condition and substitute the coefficient.

$\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^{3}}{3}+\frac{{\mathrm{t}}^{4}}{12}-\frac{{\mathrm{t}}^{5}}{24}$

Hence, the first four nonzero terms are $\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^{3}}{3}+\frac{{\mathrm{t}}^{4}}{12}-\frac{{\mathrm{t}}^{5}}{24}$. ### Want to see more solutions like these? 