Solve

$\displaystyle dy/dx + y^2 = 1$

using a 2nd order Runge Kutta and a step size of h = 0.1

When i calculate $\displaystyle y1$ i get 0.0995 for t=0.1

When i calculate $\displaystyle y2$ i get 0.1589 for t=0.2 but the answer says 0.19703

Never mind, problem solved