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January 30, 2015

January 30, 2015

Posted by **Geminese** on Tuesday, March 13, 2012 at 2:24pm.

y'= -3x+y^2, y(0)=1

- calculus 2 -
**Steve**, Tuesday, March 13, 2012 at 4:09pmwhat's the problem? Just plug and chug:

Y_{i+1}= y_{i}+ .2*y'(x_{i},y_{i})

0: (0.0000,1.0000) -> 1.2000

1: (0.2000,1.2000) -> 1.3680

2: (0.4000,1.3680) -> 1.5023

3: (0.6000,1.5023) -> 1.5937

4: (0.8000,1.5937) -> 1.6216

5: (1.0000,1.6216) -> 1.5475

Looks like f(1) = 1.6216

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