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Q 3.6-16E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 131
Fundamentals Of Differential Equations And Boundary Value Problems

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

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Short Answer

The solution to the initial value problem dydx+yx=x3y2,y(1)=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.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the equation of approximation value

Here dydx+yx=x3y2,y(1)=3,

For, xo=1,yo=3, M = 280, h = 0.005, interval = 1,2

F=f(x,y)=x3y2-yxG=f(x+h,y+hF)=(x+0.005)3y+0.005x3y2-yx2-y+0.005x3y2-yxx+0.005

Step 2: Solve for x and y

Apply initial points x=0,y=3,h=0.005

F(1,3)=6G(1,3)=6.030x=(xo+h)y=yo+h2(F+G)x=0.005y=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.

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