Denebel
use separation of variables to solve the initial value problem.
dy/dx = (y+2)(x+2) and y=1 when x=0



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Denebel
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*oops I meant (y+5)

Denebel
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dy/dx= (y+5)(x+2) and y=1 when x=0

Roachie
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dy/(y+5) = (x+2) dx
integrate both sides

Denebel
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I did, but I don't know what to do after..
I did this?
dw:1327548662810:dw

Denebel
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Can you show me your steps?

Roachie
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ok,
take your ln ( y+5) = 1/2x^2+2x + C
and raise expotentiate to get rid of the ln
y+5 = Ce^(1/2x^2+2x )
then
y = Ce^(1/2x^2+2x )  5

Roachie
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y=6e^(.5x^2+2x) + 5

Denebel
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Why can't I plug in the numbers before putting in e?

Roachie
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Because this is how I was taught :)
lemme try it that way

Roachie
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that gives ln(6) = C... 1.7917594692280550008124773583807022727229906921830047058553

Denebel
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Oh ok. thanks :) How did you get your C to become a coefficient?

Denebel
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Referring to this: y+5 = Ce^(1/2x^2+2x )

Denebel
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Is this a rule?