assinale a unica alternativa que apresenta o valor de y(3) em face da resolução da EDO de primeira ordem y'=y2, sendo y(0) =0,3. considere h =0.30....

Utilizando o método de Runge-Kutta de quarta ordem, temos: k1 = h * f(0.3, 0.3^2) = 0.027 k2 = h * f(0.3 + h/2, (0.3 + k1/2)^2) = 0.028 k3 = h * f(0.3 + h/2, (0.3 + k2/2)^2) = 0.028 k4 = h * f(0.3 + h, (0.3 + k3)^2) = 0.029 y(0.6) = y(0.3) + (k1 + 2k2 + 2k3 + k4)/6 = 0.3 + (0.027 + 2*0.028 + 0.029)/6 = 0.327 k1 = h * f(0.6, 0.327^2) = 0.034 k2 = h * f(0.6 + h/2, (0.327 + k1/2)^2) = 0.035 k3 = h * f(0.6 + h/2, (0.327 + k2/2)^2) = 0.035 k4 = h * f(0.6 + h, (0.327 + k3)^2) = 0.036 y(0.9) = y(0.6) + (k1 + 2k2 + 2k3 + k4)/6 = 0.327 + (0.034 + 2*0.035 + 0.036)/6 = 0.364 k1 = h * f(0.9, 0.364^2) = 0.042 k2 = h * f(0.9 + h/2, (0.364 + k1/2)^2) = 0.043 k3 = h * f(0.9 + h/2, (0.364 + k2/2)^2) = 0.043 k4 = h * f(0.9 + h, (0.364 + k3)^2) = 0.044 y(1.2) = y(0.9) + (k1 + 2k2 + 2k3 + k4)/6 = 0.364 + (0.042 + 2*0.043 + 0.044)/6 = 0.408 k1 = h * f(1.2, 0.408^2) = 0.054 k2 = h * f(1.2 + h/2, (0.408 + k1/2)^2) = 0.055 k3 = h * f(1.2 + h/2, (0.408 + k2/2)^2) = 0.055 k4 = h * f(1.2 + h, (0.408 + k3)^2) = 0.056 y(1.5) = y(1.2) + (k1 + 2k2 + 2k3 + k4)/6 = 0.408 + (0.054 + 2*0.055 + 0.056)/6 = 0.462 k1 = h * f(1.5, 0.462^2) = 0.072 k2 = h * f(1.5 + h/2, (0.462 + k1/2)^2) = 0.073 k3 = h * f(1.5 + h/2, (0.462 + k2/2)^2) = 0.073 k4 = h * f(1.5 + h, (0.462 + k3)^2) = 0.074 y(1.8) = y(1.5) + (k1 + 2k2 + 2k3 + k4)/6 = 0.462 + (0.072 + 2*0.073 + 0.074)/6 = 0.522 k1 = h * f(1.8, 0.522^2) = 0.097 k2 = h * f(1.8 + h/2, (0.522 + k1/2)^2) = 0.098 k3 = h * f(1.8 + h/2, (0.522 + k2/2)^2) = 0.098 k4 = h * f(1.8 + h, (0.522 + k3)^2) = 0.099 y(2.1) = y(1.8) + (k1 + 2k2 + 2k3 + k4)/6 = 0.522 + (0.097 + 2*0.098 + 0.099)/6 = 0.590 k1 = h * f(2.1, 0.590^2) = 0.146 k2 = h * f(2.1 + h/2, (0.590 + k1/2)^2) = 0.147 k3 = h * f(2.1 + h/2, (0.590 + k2/2)^2) = 0.147 k4 = h * f(2.1 + h, (0.590 + k3)^2) = 0.148 y(2.4) = y(2.1) + (k1 + 2k2 + 2k3 + k4)/6 = 0.590 + (0.146 + 2*0.147 + 0.148)/6 = 0.674 k1 = h * f(2.4, 0.674^2) = 0.246 k2 = h * f(2.4 + h/2, (0.674 + k1/2)^2) = 0.247 k3 = h * f(2.4 + h/2, (0.674 + k2/2)^2) = 0.247 k4 = h * f(2.4 + h, (0.674 + k3)^2) = 0.248 y(2.7) = y(2.4) + (k1 + 2k2 + 2k3 + k4)/6 = 0.674 + (0.246 + 2*0.247 + 0.248)/6 = 0.784 k1 = h * f(2.7, 0.784^2) = 0.386 k2 = h * f(2.7 + h/2, (0.784 + k1/2)^2) = 0.387 k3 = h * f(2.7 + h/2, (0.784 + k2/2)^2) = 0.387 k4 = h * f(2.7 + h, (0.784 + k3)^2) = 0.388 y(3) = y(2.7) + (k1 + 2k2 + 2k3 + k4)/6 = 0.784 + (0.386 + 2*0.387 + 0.388)/6 = 0.938 Portanto, a alternativa correta é a letra D) 0,938.