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Denebel
 3 years ago
use separation of variables to solve the initial value problem.
dy/dx = (y+2)(x+2) and y=1 when x=0
Denebel
 3 years ago
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
 3 years ago
Best ResponseYou've already chosen the best response.0dy/dx= (y+5)(x+2) and y=1 when x=0

Roachie
 3 years ago
Best ResponseYou've already chosen the best response.1dy/(y+5) = (x+2) dx integrate both sides

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0I did, but I don't know what to do after.. I did this? dw:1327548662810:dw

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Can you show me your steps?

Roachie
 3 years ago
Best ResponseYou've already chosen the best response.1ok, 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

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Why can't I plug in the numbers before putting in e?

Roachie
 3 years ago
Best ResponseYou've already chosen the best response.1Because this is how I was taught :) lemme try it that way

Roachie
 3 years ago
Best ResponseYou've already chosen the best response.1that gives ln(6) = C... 1.7917594692280550008124773583807022727229906921830047058553

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Oh ok. thanks :) How did you get your C to become a coefficient?

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Referring to this: y+5 = Ce^(1/2x^2+2x )
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