# Thread: Find the general solution of the differential equation specified..

1. ## Find the general solution of the differential equation specified..

Hey all,

Given the differential equation:

$\frac{dy}{dt}=\frac{1}{2y+1}$, where $x$ is not $-\frac{1}{2}$

I do not understand how they went from:
$\int(2y+1)dy = \int tdt$
$y^2+y = t + k$

to

$y(t) = \frac{-1\pm \sqrt{4t+4k+1}}{2}$

2. They completed the square on the $\displaystyle y$ terms and solved for $\displaystyle y$.

3. Originally Posted by Oiler
Hey all,

Given the differential equation:

$\frac{dy}{dt}=\frac{1}{2y+1}$, where $x$ is not $-\frac{1}{2}$

I do not understand how they went from:
$\int(2y+1)dy = \int tdt$
This should be

$\displaystyle\int(2y+1)\,dy=\int dt.$

$y^2+y = t + k$

to

$y(t) = \frac{-1\pm \sqrt{4t+4k+1}}{2}$