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If a solution curve, y, passes through the point (.5, 1.5), what is the approximate value of y(.6)?

If a solution curve, y, passes through the point (.5, 1.5), what is the approximate value of y(.6)?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Differential Equations Slope Fields. From the given table, what would be the approximate value of y(0.6)?

If a solution curve, \( y \), passes through the point (0.5, 1.5), what is the approximate value of \( y(0.6) \)?
Transcribed Image Text:If a solution curve, \( y \), passes through the point (0.5, 1.5), what is the approximate value of \( y(0.6) \)?
To get started, suppose \(\frac{dy}{dt} = \frac{t}{y}\), in which case \(f(t, y) = \frac{t}{y}\). Complete the following table of values, labeling any undefined entries with an "X". \[ \begin{array}{c|ccccccc} y \backslash t & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 \\ \hline 2 & -1 & -0.75 & -\frac{1}{2} & -\frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} & 1 \\ 1.5 & -\frac{4}{3} & -1 & -\frac{2}{3} & -\frac{1}{3} & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{4}{3} \\ 1 & -2 & -\frac{3}{2} & -1 & -\frac{1}{2} & 0 & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ 0.5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ 0 & X & X & X & X & X & X & X & X & X \\ -0.5 & 4 & 3 & 2 & 1 & 0 & -1 & -2 & -3 & -4 \\ -1 & 2 & \frac{3}{2} & 1 & \frac{1}{2} & 0 & -\frac{1}{2} & -1 & -\frac{3}{2} & -2 \\ -1.5 & \frac{4}{3} & \frac{3}{2} & \frac{2}{3} & \frac{1}{3} & 0 & -\frac{1}{3} & -\frac{2}{3} & -1 & -\frac{4}{3} \\ -
Transcribed Image Text:To get started, suppose \(\frac{dy}{dt} = \frac{t}{y}\), in which case \(f(t, y) = \frac{t}{y}\). Complete the following table of values, labeling any undefined entries with an "X". \[ \begin{array}{c|ccccccc} y \backslash t & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 \\ \hline 2 & -1 & -0.75 & -\frac{1}{2} & -\frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} & 1 \\ 1.5 & -\frac{4}{3} & -1 & -\frac{2}{3} & -\frac{1}{3} & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{4}{3} \\ 1 & -2 & -\frac{3}{2} & -1 & -\frac{1}{2} & 0 & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ 0.5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ 0 & X & X & X & X & X & X & X & X & X \\ -0.5 & 4 & 3 & 2 & 1 & 0 & -1 & -2 & -3 & -4 \\ -1 & 2 & \frac{3}{2} & 1 & \frac{1}{2} & 0 & -\frac{1}{2} & -1 & -\frac{3}{2} & -2 \\ -1.5 & \frac{4}{3} & \frac{3}{2} & \frac{2}{3} & \frac{1}{3} & 0 & -\frac{1}{3} & -\frac{2}{3} & -1 & -\frac{4}{3} \\ -
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Approximate by straightline and find slope using table.

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