x = 0.3, and y0 = 0 so, interval ==>[0,0.3] step soize = h = 0.1 applying Euler approximation x(n+1) = xn + h and y(n+1) = yn + hf(xn,yn) x(n+1) = xn + h x1 = x0 + h = 0.1 y(n+1) = yn + hf(xn,yn) or, y1 = y0 + hf(x0,y0) = 1 + 0.1(xo+yo) = 1 + 0.1(0+1) = 1.1 so, x1 = 0.1 y1 = 1.1 again x(n+1) = xn + h x2 = x1 + h = 0.2 y(n+1) = yn + hf(xn,yn) or, y2 = y1 + hf(x1,y1) = 1.1 + 0.1*(x1 + y1) = 1.1 + 0.1(0.1 + 1.1) = 1.22 so, x2 = 0.2 y2 = 1.22 again x(n+1) = xn + h x3 = x2 + h = 0.3 y(n+1) = yn + hf(xn,yn) or, y3 = y2 + hf(x2,y2) = 1.22 + 0.1*(x2 + y2) = 1.22 + 0.1(0.2 + 1.22) = 1.363 so, x3 = 0.3 y3 = 1.362 again x(n+1) = xn + h x4 = x3 + h = 0.4 y(n+1) = yn + hf(xn,yn) or, y4 = y3 + hf(x3,y) = 1.362 + 0.1*(x3 + y3) = 1.362 + 0.1(0.3 + 1.362) = 1.5282 so, x2 = 0.4 y2 = 1.5282 the graph of the data set generated will yield the approximate graph of the original equation Solution x = 0.3, and y0 = 0 so, interval ==>[0,0.3] step soize = h = 0.1 applying Euler approximation x(n+1) = xn + h and y(n+1) = yn + hf(xn,yn) x(n+1) = xn + h x1 = x0 + h = 0.1 y(n+1) = yn + hf(xn,yn) or, y1 = y0 + hf(x0,y0) = 1 + 0.1(xo+yo) = 1 + 0.1(0+1) = 1.1 so, x1 = 0.1 y1 = 1.1 again x(n+1) = xn + h x2 = x1 + h = 0.2 y(n+1) = yn + hf(xn,yn) or, y2 = y1 + hf(x1,y1) = 1.1 + 0.1*(x1 + y1) = 1.1 + 0.1(0.1 + 1.1) = 1.22 so, x2 = 0.2 y2 = 1.22 again x(n+1) = xn + h x3 = x2 + h = 0.3 y(n+1) = yn + hf(xn,yn) or, y3 = y2 + hf(x2,y2) = 1.22 + 0.1*(x2 + y2) = 1.22 + 0.1(0.2 + 1.22) = 1.363 so, x3 = 0.3 y3 = 1.362 again x(n+1) = xn + h x4 = x3 + h = 0.4 y(n+1) = yn + hf(xn,yn) or, y4 = y3 + hf(x3,y) = 1.362 + 0.1*(x3 + y3) = 1.362 + 0.1(0.3 + 1.362) = 1.5282 so, x2 = 0.4 y2 = 1.5282 the graph of the data set generated will yield the approximate graph of the original equation.