Use Euler's method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05.
Find an explicit solution for the initial-value problem and then fill in the following tables. (Round your answers to four decimal places. Percentages may be rounded to two decimal places.) y' = 2xy, y(1) = 1; y(1.5) y(x) = (explicit solution)
h = 0.1
xn yn Actual Value Absolute Error % Rel. Error
1.00 1.0000 1.0000 0.0000 0.00
1.10 1.2337 1.20 1.5527 1.30 1.9937
1.40 2.6117 1.50 3.4903
h = 0.05
xn yn Actual Value Absolute Error % Rel. Error
1.00 1.0000 1.0000 0.0000 0.00
1.05 1.1000 1.1079 0.0079 0.71
1.10 1.2155 1.2337 0.0182 1.48
1.15 1.3492 1.3806 0.0314 2.27
1.20 1.5044 1.5527 0.0483 3.11
1.25 1.6849 1.7551 0.0702 4.00
1.30 1.8955 1.9937 0.0982 4.93
1.35 2.1419 2.2762 0.1343 5.90
1.40 2.4311 2.6117 0.1806 6.92
1.45 2.7715 3.0117 0.2402 7.98

[tex]\displaystyle\dfrac{dy}{dx}=2xy\\\\\int{\dfrac{dy}{y}}=\int{2x}\,dx\\\\\ln{y}=x^2+C\\\\\text{Considering the initial condition, $C=-1$}\\\\\boxed{y=e^{x^2-1}}[/tex]

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The values for yn are y+y'·h = y+2xyh. We take the "absolute error" to be the (signed) difference between the calculated yn and the actual value y(x).

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