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2 answers

well it goes like this

y^-2dy = dx/(x+5)

integrate both sides and finished

so you get
-1/y = ln(x+5) + C

C is a constant that depends on some initial value that you didnt give.

2006-07-26 08:57:42 · answer #1 · answered by gjmb1960 7 · 0 0

Ooh. Diffy Q's. I love diffy Q's!

Separation of variables just means: put everything that involves x next to the dx, everything that involves y next to the dy, and integrate both sides.

So we start with
(x+5) dy/dx = y^2
Move the dx to the other side:
(x+5) dy = y^2 dx
This is close to what we need, but everything's on the wrong side. So divide by (x+5) and by y^2.
1/y^2 dy = 1/(x+5) dx
Slap integral signs on both sides:
S 1/y^2 dy = S 1/(x+5) dx
Integrate:
-1/y = ln |x+5| + C

This is a valid answer right here. However, most of the time you'll want to find out what C is (if you have a boundary condition specified? maybe? no? ok then), and you'll want to solve for the format y=something. which isn't too hard.

2006-07-26 08:58:32 · answer #2 · answered by geofft 3 · 0 0

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