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

Mathematics
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*oops I meant (y+5)
dy/dx= (y+5)(x+2) and y=1 when x=0
dy/(y+5) = (x+2) dx integrate both sides

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I did, but I don't know what to do after.. I did this? |dw:1327548662810:dw|
Can you show me your steps?
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
y=6e^(.5x^2+2x) + 5
Why can't I plug in the numbers before putting in e?
Because this is how I was taught :) lemme try it that way
that gives ln(6) = C... 1.7917594692280550008124773583807022727229906921830047058553
Oh ok. thanks :) How did you get your C to become a coefficient?
Referring to this: y+5 = Ce^(1/2x^2+2x )
Is this a rule?

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