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Q 3.6-16E

Expert-verifiedFound in: Page 131

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.**

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

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

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

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

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

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.**

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