Use the improved Euler method to find approximate values of the solution of the initial-value problem y'=2x^2+3y^2-2, y(2)=1; h=0.05 at the points Xi=X0+ih, where X0 is the point where the initial condition is imposed and i=1, 2, 3. sd

Question

idealist10 · Answer

@SithsAndGiggles @ganeshie8 @wio @Zarkon

anonymous · Answer

Hmmm

anonymous · Answer

Are we to use $y(2)=1$ as the initial condition and work from there?  And where are we going to?

anonymous · Answer

I dunno, I guess this is what they want you to do.

anonymous · Answer

So we start with $y(2)=1$.Then: $$
y(2+0.05)=y(2.05)\approx 1 + 0.05\bigg(2(2)^2+3(1)^2-2\bigg)
$$

anonymous · Answer

This will give you that $y(2.05)=\ldots$ which you'll use in the next approximation.

anonymous · Answer

$$
y(2.1) \approx y(2.05)+0.05\bigg(2(2.05)^2+3[y(2.05)]^2-2\bigg)
$$

anonymous · Answer

And I assume you do this three times?