Hallar los valores aproximados de la solucion del problema dado con valorinicial, en t = t0 + 0.1, t0 + 0.2, t0 + 0.3 y t0 + 0.4:

a) Aplicar el metodo de Euler con h = 0.1

b) Aplicar el metodo de Euler con h = 0.05

c) Aplicar el metodo de Euler con h = 0.025

1. y = ex2, x(0) = 1

a) Aplicando el metodo de Euler: yn+1 = yn + hf(tn, yn) con h = 0.1Sea t1 = 1 + 0.1Para t1 = 1.1y1 = 3 + (0.1)f(1.1 , 3)y1 = 3 + (0.1)(2.0248)y1 = 3.2024

Para t2 = 1.2y2 = 3.2024 + (0.1)f(1.2 , 3.2024)y2 = 3.2024 + (0.1)(2.0982)y2 = 3.4122

Para t3 = 1.3y3 = 3.4122 + (0.1)f(1.3 , 3.4122)y3 = 3.4122 + (0.1)(2.1707)y3 = 3.6292

Para t4 = 1.4y4 = 3.6292 + (0.1)f(1.2 , 3.2024)y4 = 3.6292 + (0.1)(2.0982)y4 = 3.4122

b) Aplicando el metodo de Euler: yn+1 = yn + hf(tn, yn) con h = 0.05

Para t1 = 1.1y1 = 3 + (0.05)f(1.1 , 3)y1 = 3 + (0.05)(2.0248)y1 = 3.1012

Para t2 = 1.15y2 = 3.1012 + (0.05)f(1.15 , 3.1012)

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y2 = 3.1012 + (0.05)(2.0618)y2 = 3.2042

Para t3 = 1.2y3 = 3.2042 + (0.05)f(1.2 , 3.2042)y3 = 3.2042 + (0.05)(2.0986)y3 = 3.3091

Para t4 = 1.25y4 = 3.3091 + (0.05)f(1.25 , 3.3091)y4 = 3.3091 + (0.05)(2.1352)y4 = 3.4158

Para t5 = 1.3y5 = 3.4158 + (0.05)f(1.3 , 3.4158)y5 = 3.4158 + (0.05)(2.1716)y5 = 3.5243

Para t6 = 1.35y6 = 3.5243 + (0.05)f(1.35 , 3.5243)y6 = 3.5243 + (0.05)(2.2077)y6 = 3.6346

Para t7 = 1.4y7 = 3.6346 + (0.05)f(1.4 , 3.6346)y7 = 3.6346 + (0.05)(2.2438)y7 = 3.7467

Para t8 = 1.45y8 = 3 + (0.05)f(1.45 , 3.7467)y8 = 3 + (0.05)(2.2796)y8 = 3.860

c) De manera similar se obtienen los resultados de h = 0.025 y se muestranen la siguiente tabla:

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n tn yn f(tn, yn) Aproximacion1 1.025 3 2.1246 3.05032 1.05 3.0503 2.0310 3.10103 1.075 3.1010 2.0496 3.15234 1.1 3.1523 2.0681 3.20405 1.125 3.2040 2.0866 3.25626 1.15 3.2562 2.1050 3.30887 1.175 3.3088 2.1233 3.36198 1.2 3.3619 2.1417 3.41549 1.225 3.4154 2.1599 3.469410 1.25 3.4694 2.1781 3.523911 1.275 3.5239 2.1963 3.578812 1.3 3.5788 2.2144 3.634113 1.325 3.6368 2.2325 3.689914 1.35 3.6899 2.2505 3.746215 1.375 3.7462 2.2685 3.802916 1.4 3.8029 2.2864 3.8601

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