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Answers without the blur. Sign up and see all textbooks for free! Q 3.7-6E

Expert-verified Found in: Page 139 ### 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 # Use the Taylor methods of orders 2 and 4 with h = 0.25 to approximate the solution to the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$, at x = 1. Compare these approximations to the actual solution ${\mathbf{y}}{\mathbf{=}}{\mathbf{1}}{\mathbf{+}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{x}}$ evaluated at x = 1.

${\mathbit{\varphi }}_{\mathbf{2}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6274}\phantom{\rule{0ex}{0ex}}{\mathbit{\varphi }}_{\mathbf{4}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6227}$

See the step by step solution

## Step 1: Find the value of f2(x,y)

Here $\mathrm{y}\text{'}=1-\mathrm{y},\mathrm{y}\left(0\right)=0$

Apply the chain rule.

${\mathrm{f}}_{2}\left(\mathrm{x},\mathrm{y}\right)=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\left(\mathrm{x},\mathrm{y}\right)+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$

Since ${\mathrm{f}}_{2}\left(\mathrm{x},\mathrm{y}\right)=1-y$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\left(\mathrm{x},\mathrm{y}\right)=0\phantom{\rule{0ex}{0ex}}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\left(\mathrm{x},\mathrm{y}\right)=-1$

So, the equation is ${\mathrm{f}}_{2}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{y}-1$

## Step 2: Evaluate the values of f3(x,y) and f4(x,y)

Apply the same procedure as step 1

${\mathrm{f}}_{3}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{y}-1\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{4}\left(\mathrm{x},\mathrm{y}\right)=1-\mathrm{y}$

## Step 3: Apply the recursive formulas for order 2

The recursive formula is

## Step 4: Apply the initial condition and find the values of x and y

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_{0}=0$

${\mathrm{x}}_{1}=0.25\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{1}=0.21875\phantom{\rule{0ex}{0ex}}$

Put all these values in recursive formulas for the other values.

${\mathrm{x}}_{2}=0.5\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{2}=0.389648\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}=0.75\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{3}=0.523163\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{4}=1\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{4}=0.627471$

Therefore the approximation of the solution by the Taylor method of order 2 at point x=1

${\varphi }_{2}\left(1\right)=0.6274$

## Step 6: Apply the initial condition and find the values of x and y

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_{0}=0$.

${\mathrm{x}}_{1}=0.25\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{1}=0.216309$

Put all these values in recursive formulas for the other values.

${\mathrm{x}}_{2}=0.5\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{2}=0.385828\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{3}=0.75\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{3}=0.518679\phantom{\rule{0ex}{0ex}}{\mathrm{x}}_{4}=1\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{4}=0.622793$

Thus, the approximation of the solution by the Taylor method of order 4 at point x = 1

${\varphi }_{4}\left(1\right)=0.6227$

The actual solution at x = 1

$\mathrm{y}\left(\mathrm{x}\right)=1-{\mathrm{e}}^{-\mathrm{x}}\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(1\right)=0.632121$

Now combining the approximation with the actual solution at x = 1

$\left|\mathrm{y}\left(1\right)-{\varphi }_{2}\left(1\right)\right|=0.00465\phantom{\rule{0ex}{0ex}}\left|\mathrm{y}\left(1\right)-{\varphi }_{4}\left(1\right)\right|=0.00933\phantom{\rule{0ex}{0ex}}$

Hence the solution is

${\mathbit{\varphi }}_{\mathbf{2}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6274}\phantom{\rule{0ex}{0ex}}{\mathbit{\varphi }}_{\mathbf{4}}\left(1\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6227}$ ### Want to see more solutions like these? 