Chapter 3: Q 3.6-16E (page 131)

The solution to the initial value problem $\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}, y (1) = 3$ 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.

Short Answer

Expert verified

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

Step by step solution

Find the equation of approximation value

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

For, $x_{o} = 1, y_{o} = 3, M = 280, h = 0.005, interval = [1, 2]$

$F = f (x, y) = (x^{3} y^{2} - \frac{y}{x}) G = f (x + h, y + hF) = (x + 0 . 005)^{3} {(y + 0.005 (x^{3} y^{2} - \frac{y}{x}))}^{2} - \frac{y + 0.005 (x^{3} y^{2} - \frac{y}{x})}{x + 0.005}$

Solve for x and y

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

$F (1, 3) = 6 G (1, 3) = 6 . 030 x = (x_{o} + h) y = y_{o} + \frac{h}{2} (F + G) x = 0.005 y = 3$

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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The solution to the initial value problem $\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}, y (1) = 3$ 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.

Short Answer

Step by step solution

Find the equation of approximation value

Solve for x and y

Determine the all-other values

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The solution to the initial value problemdydx+yx=x3y2,y(1)=3has 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.

Short Answer

Step by step solution

Find the equation of approximation value

Solve for x and y

Determine the all-other values

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Statistics

Calculus

Mechanics Maths

Theoretical and Mathematical Physics

Pure Maths

Study anywhere. Anytime. Across all devices.

The solution to the initial value problem $\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}, y (1) = 3$ 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.