Originally Posted by

**kpizle** Hey.

My problem is $\displaystyle y(t)' + (1/y(t)+1) = 0$, where $\displaystyle y(0) =2$.

I am having a hard time getting y(t).

At first glance, this looks like a homogenous ODE...but using the form $\displaystyle y'(t) + p(t)y(t)=0$, what would $\displaystyle p(t)$ be?

After hitting that dead end, I decided to try moving the $\displaystyle (1/y(t)+1)$ to the right side and doing

$\displaystyle y'(t) = -(1/y(t)+1)$ then

$\displaystyle dy/dt = -(1/y(t)+1)$ then

$\displaystyle dy = -(1/y(t)+1) dt$ and acting like $\displaystyle -(1/y(t)+1)$ is a constant.

When I try to do this, I end up with $\displaystyle y = (-t/(y(t)+1)) + C$

which leaves me with $\displaystyle y(t)^2 + y(t) = t$...and I get stuck.

Also, I have tried getting $\displaystyle (y(t)+2)/(y(t) + 1)$ on the left side by making the left side one fraction and moving the dt over to the right...after I integrate, I get $\displaystyle ln(|y(t)+1|)+y(t)=C$...and I get stuck.

Am I forgetting some algebra here, or am I completely lost?

Please help me!

Thanks!