ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: liv on June 21, 2010, 09:30:23 pm
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can anyone explain this? i don't get why it is used.
and this may be a simple question, but how do you do:
using euler's method, approximate y(2) with i) h=0.5 and ii)0.2 for dy/dx=x/(sqrt(x^2+8), y(1)=3
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Euler's Method: yn+1 = yn + h*f'(xn)
So in this case, with h=0.5, y2 = 3 + 0.5*1/3
That gives y2 = 19/6, repeat that with the new values to find y3, which is an approximation for y(2).
It's the same with steps of 0.2, it will just take 5 applications of Euler's method, each building on the last, until you've found y6.