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

• Mar 13th 2011, 10:57 PM
Oiler
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

$y(t) = \frac{-1\pm \sqrt{4t+4k+1}}{2}$
• Mar 13th 2011, 11:44 PM
Prove It
They completed the square on the $\displaystyle y$ terms and solved for $\displaystyle y$.
• Mar 14th 2011, 02:39 AM
Ackbeet
Quote:

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

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

Quote:

$y^2+y = t + k$

to

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