Dado que open curly brackets table attributes columnalign left end attributes row cell fraction numerator d y over denominator d x end fraction eq...

Dado que open curly brackets table attributes columnalign left end attributes row cell fraction numerator d y over denominator d x end fraction equals negative y plus x plus 2 end cell row cell y left parenthesis 0 right parenthesis equals 2 end cell end table close . Usando o método de Runge-Kutta de 4ª ordem com x element of left square bracket 0 semicolon 0 comma 3 right square bracket e h equals 0 comma 1, encontra-se que y left parenthesis 0 comma 3 right parenthesis é, aproximadamente, igual a: