Denebel
What happens after this..?
use separation of variables to solve the initial value problem



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Denebel
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Denebel
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Forgot the parenthesis: e^( (x^2/2) + 2x + c)

Roachie
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sub youe initial values and solve for C
y= C(e^( .5x^2 + 2x )) 5
for x = 0, y = 1
1 = C  5
C = 6
so
y= 6e^( .5x^2 + 2x ) 5

cinar
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ln(6)=c

Denebel
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How did you bring C out?

Roachie
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e^C = C
its math magic

Denebel
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? Huh?
But for example, e^(2) does not equal to 2...

cinar
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dw:1327563797536:dw

Roachie
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e^c = 6

cinar
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dw:1327563871600:dw

Roachie
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cinar!

cinar
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yes!

Denebel
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ahh I see now! Thank you very much

Roachie
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just try to have fun with it.