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Expert-verified Found in: Page 426 ### 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 # The solution to the initial value problem$\begin{array}{l}x{y}^{\text{'}\text{'}}\left(x\right)+2{y}^{\text{'}}\left(x\right)+xy\left(x\right)=0\\ y\left(0\right)=1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{y}^{\text{'}}\left(0\right)=0\end{array}$has derivatives of all orders at ${\mathbit{x}}{\mathbf{=}}{\mathbf{0}}$ (although this is far from obvious). Use L'Hôpital's rule to compute the Taylor polynomial of degree 2 approximating this solution.

The required polynomial is,${p}_{3}\left(x\right)=1-\frac{{x}^{2}}{6}$ .

See the step by step solution

## Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at ,${x}_{0}$ approximating a function$f\left(x\right)$ possessing n derivatives at , ${x}_{0}$is given by

${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\text{'}}\left({x}_{0}\right)×\left(x-{x}_{0}\right)+{f}^{\text{'}\text{'}}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+\cdots +{f}^{n}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{n}}{n!}$

The differential equation is given as

$x{y}^{\text{'}\text{'}}\left(x\right)+2{y}^{\text{'}}\left(x\right)+xy\left(x\right)=0$

It is given that for the function ,$y\left(x\right)$

$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

For applying the L’Hospital rule we need to see that our differential equation satisfies $\frac{0}{0}$form,

$\begin{array}{c}x{y}^{\text{'}\text{'}}\left(x\right)+2{y}^{\text{'}}\left(x\right)+xy\left(x\right)=0\\ {y}^{\text{'}\text{'}}\left(x\right)+\frac{2{y}^{\text{'}}\left(x\right)}{x}+y\left(x\right)=0\\ {y}^{\text{'}\text{'}}\left(x\right)=-y\left(x\right)-\frac{2{y}^{\text{'}}\left(x\right)}{x}\\ {y}^{\text{'}\text{'}}\left(x\right)=\frac{-xy\left(x\right)-2{y}^{\text{'}}\left(x\right)}{x}\\ {y}^{\text{'}\text{'}}\left(0\right)=0\end{array}$

Hence, it satisfies L'Hospital rule and by applying it in differential equation it becomes,

$\begin{array}{c}{y}^{\text{'}\text{'}}\left(x\right)=\frac{-xy\left(x\right)-2{y}^{\text{'}}\left(x\right)}{x}\\ {y}^{\text{'}\text{'}}\left(x\right)=-2{y}^{\text{'}\text{'}}\left(x\right)-y\left(x\right)-x{y}^{\text{'}}\left(x\right)\\ {y}^{\text{'}\text{'}}\left(0\right)=-2{y}^{\text{'}\text{'}}\left(0\right)-y\left(0\right)-x{y}^{\text{'}}\left(0\right)\\ {y}^{\text{'}\text{'}}\left(0\right)=\frac{-1}{3}\end{array}$

## Step 2:Final proof

The Taylor polynomial of degree 2 in the solution is given by.${p}_{3}\left(x\right)=1-\frac{{x}^{2}}{6}$ ### Want to see more solutions like these? 