Wikipedia:Reference desk/Archives/Mathematics/2014 February 4

= February 4 =

Derivative question
I have a problem that says:

$${dy \over dx}+1 = 2y$$

I've gone through the following:

$$dy = 2y-1dx$$

$$\int dy - 2y = \int -dx$$

$$-y^2 = -x$$

This is not what my teacher got. Where have I gone wrong? It's something to do with moving the dx over to the right side, isn't it? It should be $$(2y-1)dx$$ shouldn't it? Yes, this is basic. But my answer works out for the rest of the problem, so it's messing with my brain at this hour. Thanks, Dismas |(talk) 10:06, 4 February 2014 (UTC)
 * Yes it should be $$dy = (2y-1)dx$$. And then try getting all the y's on the left, division is your friend here. Plus you always get a constant added after integration. Dmcq (talk) 10:22, 4 February 2014 (UTC)

dy/dx+1=2y

dy=(2y−1)dx

d(2y−1)=2dy=2(2y−1)dx

d(2y−1)/(2y−1)=2dx

d(log(2y−1))=d(2x)

log(2y−1)=2x+K

2y−1=e2x+K=e2xeK=Ae2x

y=(1+Ae2x)/2

Bo Jacoby (talk) 14:26, 4 February 2014 (UTC).