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Q14 E

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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 nonzero terms in a power series expansion of the solution to the given initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$

The first four nonzero terms in the power series expansion of the given initial value problem $\mathrm{y}\text{'}-{\mathrm{e}}^{\mathrm{x}}\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{x}+{\mathrm{x}}^{2}+\frac{5}{6}{\mathrm{x}}^{3}+\cdots$.

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 expression after expansion.

Given,

$\mathrm{y}\text{'}-{\mathrm{e}}^{\mathrm{x}}\mathrm{y}=0;\mathrm{y}\left(0\right)=1$

From the above equation put $\mathrm{p}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$ which is analytic over the entire number line.

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,

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

The series expansion for the function is

${\mathrm{e}}^{\mathrm{x}}=1+\mathrm{x}+\frac{{\mathrm{x}}^{2}}{2!}+\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{4}}{4!}+\cdots$

By expanding the series we get,

$\left({\mathrm{a}}_{1}+2{\mathrm{a}}_{2}\mathrm{x}+3{\mathrm{a}}_{3}{\mathrm{x}}^{2}+4{\mathrm{a}}_{4}{\mathrm{x}}^{3}\cdots \right)-\left(1+\mathrm{x}+\frac{{\mathrm{x}}^{2}}{2!}+\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{4}}{4!}+\cdots \right)\left({\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{a}}_{2}{\mathrm{x}}^{2}+{\mathrm{a}}_{3}{\mathrm{x}}^{3}+\cdots \right)=0$

Hence the expression after the expansion is:

$\left({\mathrm{a}}_{1}+2{\mathrm{a}}_{2}\mathrm{x}+3{\mathrm{a}}_{3}{\mathrm{x}}^{2}+4{\mathrm{a}}_{4}{\mathrm{x}}^{3}\cdots \right)-\left(1+\mathrm{x}+\frac{{\mathrm{x}}^{2}}{2!}+\frac{{\mathrm{x}}^{3}}{3!}+\frac{{\mathrm{x}}^{4}}{4!}+\cdots \right)\left({\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{x}+{\mathrm{a}}_{2}{\mathrm{x}}^{2}+{\mathrm{a}}_{3}{\mathrm{x}}^{3}+\cdots \right)=0$

## Step 3: Find the first four nonzero terms.

Expand the expression given in the previous step.

$\left({a}_{1}+2{a}_{2}x+3{a}_{3}{x}^{2}+4{a}_{4}{x}^{3}+5{a}_{5}{x}^{4}+\cdots \right)+\left(-{a}_{0}-{a}_{1}x-{a}_{2}{x}^{2}-{a}_{3}{x}^{3}-{a}_{4}{x}^{4}+\cdots \right)\phantom{\rule{0ex}{0ex}}+\left(-{a}_{0}x-{a}_{1}{x}^{2}-{a}_{2}{x}^{3}-{a}_{3}{x}^{4}-{a}_{4}{x}^{5}+\cdots \right)+\left(-{a}_{0}\frac{{x}^{2}}{2}-{a}_{1}\frac{{x}^{3}}{2}-{a}_{2}\frac{{x}^{4}}{2}-{a}_{3}\frac{{x}^{5}}{2}-\cdots \right)$

$\left({\mathrm{a}}_{1}-{\mathrm{a}}_{0}\right)+\left(2{\mathrm{a}}_{2}-{\mathrm{a}}_{1}-{\mathrm{a}}_{0}\right)\mathrm{x}+\left(3{\mathrm{a}}_{3}-{\mathrm{a}}_{2}-{\mathrm{a}}_{1}-\frac{{\mathrm{a}}_{0}}{2}\right){\mathrm{x}}^{2}+\cdots =0$

By equating the coefficients,

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

The general solution was

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

Apply the initial condition and substitute the coefficient.

$\mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{x}+{\mathrm{x}}^{2}+\frac{5}{6}{\mathrm{x}}^{3}+\cdots$

Hence, the first four nonzero terms are $\mathrm{y}\left(\mathrm{x}\right)=1+\mathrm{x}+{\mathrm{x}}^{2}+\frac{5}{6}{\mathrm{x}}^{3}+\cdots$.