Question

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; (1.5) (explicit solution) h = 0.1 Actual хо Yn Value Absolute Error % Rel. Error 1.00 1.0000 1.0000 0.0000 0.00 1.2337 1.10 1.20 1.5527 1.9937 1.30 1.40 2.6117 3.4903 1.50 h = 0.05 Actual Value xn VO Absolute Error % Rel. Error 1.00 1.0000 1.0000 0.0000 1.05 1.1000 1.1079 0.0079 0.00 0.71 1.48 1.10 1.2155 1.2337 0.0182 1.3492 1.3806 0.0314 2.27 1.15 1.20 1.25 1.5044 1.5527 3.11 1.7551 0.0483 0.0702 0.0982 1.6849 1.8955 4.00 1.30 1.9937 4.93 5.90 1.35 2.1419 2.2762 0.1343 1.40 0.1806 2.4311 2.7715 2.6117 3.0117 6.92 7.98 1.45 0.2402 1.50 3.4903

Answer 1

The IVP is

Solve the above IVP by seperation of variables

$\frac{dy}{dx}=2xy => \frac{dy}{y}=2xdx => lny=\frac{2x^2}{2}+C => lny=x^2+C$

Calculate C by applying the initial conditions

Therefore, the solution of the IVP is

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From Eulers Method, the iterations for the solution are given as

For the given IVP,

Therefore, for h=0.1 implies

$y_{2}=y_{1}+0.1f(x_{1},y_{1})=1.2+0.1(2)(1.1)(1.2)=1.464$

The rest of the iterations and the errors are shown below

From above,

$y(1.5)\approx 2.9278$

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Therefore, for h=0.05 implies

$y_{2}=y_{1}+0.05f(x_{1},y_{1})=1.1+0.05(2)(1.05)(1.1)=1.2155$

The rest of the iterations and the errors are shown in below

From above,

$y(1.5)\approx 3.1733$

Answer 2

1 ac 3 y 0 = 1 + c =) --| lny = 2², 14 - 20dx, integrating long 962-17 Into dont hf(xo, Yp), f(x,y) = 2ay Yo x² f. Rel. Error an Actual valu Abrohte error 1 1.0000 1.UDDD 1+1 X 2X7X/ = 1.2000 1-2337 0337 2.73 1.5527 .0887 5 71 8-94 1.9937 . 1783 1.2 1.2+ .1822 1.1x1.2 1-4640 1-3 1.464 + •1X2X/• 2x 1.464 1.8154 1.4 1.8154 + 2 X 2xl-3x 1.8154 2.32 37 los 2.3237 +- 1X2X1:48 2.3237 11.03 2.6117 .288 .516 14.78 2.9743 3.4903 h = .05 Acheal valu Absolute error t. Rol Error xn Yn 1.50 9.08 -3169 3.4903 2.7715 to5x2x1.450 2.7715 - 3.17 34