process of runge kutta method
to estimate (0.5) with h=0.25 dy/dx=y-x^2 & y(0)=0.1

Question

process of runge kutta method
to estimate (0.5) with h=0.25 dy/dx=y-x^2 & y(0)=0.1

anonymous · Answer

you mean y(0.5)?

anonymous · Answer

yep

anonymous · Answer

what order?

anonymous · Answer

4th order
sorry my friend had arrived

anonymous · Answer

required:
$$h=.25\quad x_0=0\quad f(x_n)={dy\over dx}=y(x_n)-x_n^2\quad x_N=5\
N=\frac{x_N-x_0}{h}
$$
then interationf from n=1 to n=N

anonymous · Answer

4th order, you have k1, k2, k3, k4
$$k_1=f(x_n,y_n)\
k_2=f\left(x_n+{h\over2},y_n+{hk_1\over2}\right)\
k_3=f\left(x_n+{h\over2},y_n+{hk_2\over2}\right)\
k_4=f(x_n+k,y_n+hk_3)
$$
and
$$
y_{n+1}=y_n+{h\over6}(k_1+2k_2+2k_3+k_4)
$$

anonymous · Answer

start with n=0 to find y_1
for that, you first find the four k and find y_1

then n=1, again find the four k, and find y2

keep going till you get to n=N-1

anonymous · Answer

oh i see
thanks man,you explain this stuff in easy way :)

ajprincess · Answer

isnt t
$k1=hf(x_n,y_n)$ this?

anonymous · Answer

oh
yeah 
just checked
:)
hf()

anonymous · Answer

I put "h" in evaluating "y" not "k"
programatically, this way is efficient for computations

anonymous · Answer

oh