Use Euler's method with h=0.05 to find approximate values for the solution of the initial-value problem y'=2x^2+3y^2-2, y(2)=1 at x=0.1, 0.2, 0.3.

Question

idealist10 · Answer

@SithsAndGiggles 
f(x, y)=2x^2+3y^2-2
$$y(2.05)\approx y _{1}=y _{0}+f(2, 1)(0.05)=1.45$$

idealist10 · Answer

How do I find the next ones?

anonymous · Answer

The next point would be
$$y(2.1)\approx y_2=y_1+f(2,1)(0.05)=\cdots$$
You use the same recursion for successive approximations to $y$,

idealist10 · Answer

How about the next one? When do I stop?

anonymous · Answer

The general procedure would be to stop once you've reached a certain threshold... The way your question is stated, you're not exactly given the threshold, but I think you're only asked to find an approximation to $y$ when $x=0.1,0.2,0.3$, i.e. $y(0.1),~y(0.2),~y(0.3)$.

idealist10 · Answer

I see. Today I have no time. Tomorrow I'll post a new question, I think I missed some info. Thanks for the help.

anonymous · Answer

yw