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Q 3.6-16E

Expert-verified
Found in: Page 131

### 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 $\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{+}}\frac{\mathbf{y}}{\mathbf{x}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{3}}}{{\mathbf{y}}}^{{\mathbf{2}}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}$ has a vertical asymptote (“blows up”) at some point in the interval [1,2] By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

The solution on the given conditions and on the interval (“blows up”) at x = 1.26.

See the step by step solution

## Step 1: Find the equation of approximation value

Here $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^{3}{\mathrm{y}}^{2},\mathrm{y}\left(1\right)=3$,

For, ${\mathrm{x}}_{\mathrm{o}}=1,{\mathrm{y}}_{\mathrm{o}}=3,\text{M = 280, h = 0.005, interval =}\left[1,2\right]$

$\mathrm{F}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\left({\mathrm{x}}^{3}{\mathrm{y}}^{2}-\frac{\mathrm{y}}{\mathrm{x}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{G}=\mathrm{f}\left(\mathrm{x}+\mathrm{h},\mathrm{y}+\mathrm{hF}\right)\phantom{\rule{0ex}{0ex}}=\left(\mathrm{x}+0.005{\right)}^{3}{\left(\mathrm{y}+0.005\left({\mathrm{x}}^{3}{\mathrm{y}}^{2}-\frac{\mathrm{y}}{\mathrm{x}}\right)\right)}^{2}-\frac{\mathrm{y}+0.005\left({\mathrm{x}}^{3}{\mathrm{y}}^{2}-\frac{\mathrm{y}}{\mathrm{x}}\right)}{\mathrm{x}+0.005}\phantom{\rule{0ex}{0ex}}$

## Step 2: Solve for x and y

Apply initial points $\mathrm{x}=0,\mathrm{y}=3,\mathrm{h}=0.005$

$\mathrm{F}\left(1,3\right)=6\phantom{\rule{0ex}{0ex}}\mathrm{G}\left(1,3\right)=6.030\phantom{\rule{0ex}{0ex}}\mathrm{x}=\left({\mathrm{x}}_{\mathrm{o}}+\mathrm{h}\right)\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{y}}_{\mathrm{o}}+\frac{\mathrm{h}}{2}\left(\mathrm{F}+\mathrm{G}\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}=0.005\phantom{\rule{0ex}{0ex}}\mathrm{y}=3$

## Step 3: Determine the all-other values

Apply the same procedure for all other values and the values are

(x = 1.261, y = 2197….)

(x = 1.262, y = 11800…)

Hence, the solution on the given conditions and on the interval cross x-axis at x = 1.26.