# Solving a DE...need help!

• Feb 10th 2010, 12:27 AM
kpizle
Solving a DE...need help!
Hey.

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

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

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

After hitting that dead end, I decided to try moving the $(1/y(t)+1)$ to the right side and doing
$y'(t) = -(1/y(t)+1)$ then
$dy/dt = -(1/y(t)+1)$ then
$dy = -(1/y(t)+1) dt$ and acting like $-(1/y(t)+1)$ is a constant.

When I try to do this, I end up with $y = (-t/(y(t)+1)) + C$
which leaves me with $y(t)^2 + y(t) = t$...and I get stuck.

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

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

Thanks!
• Feb 10th 2010, 12:53 AM
Prove It
Quote:

Originally Posted by kpizle
Hey.

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

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

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

After hitting that dead end, I decided to try moving the $(1/y(t)+1)$ to the right side and doing
$y'(t) = -(1/y(t)+1)$ then
$dy/dt = -(1/y(t)+1)$ then
$dy = -(1/y(t)+1) dt$ and acting like $-(1/y(t)+1)$ is a constant.

When I try to do this, I end up with $y = (-t/(y(t)+1)) + C$
which leaves me with $y(t)^2 + y(t) = t$...and I get stuck.

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

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

Thanks!

Is the DE

$y'(t) + \frac{1}{y(t) + 1} = 0$

or

$y'(t) + \frac{1}{y(t)} + 1 = 0$?
• Feb 10th 2010, 01:14 AM
kpizle
Quote:

Originally Posted by Prove It
Is the DE

$y'(t) + \frac{1}{y(t) + 1} = 0$

or

$y'(t) + \frac{1}{y(t)} + 1 = 0$?

It's $y'(t) + \frac{1}{y(t) + 1} = 0$

• Feb 10th 2010, 01:20 AM
Prove It
Quote:

Originally Posted by kpizle
It's $y'(t) + \frac{1}{y(t) + 1} = 0$

$y'(t) + \frac{1}{y(t) + 1} = 0$

$\frac{dy}{dt} = -\frac{1}{y + 1}$

$\frac{dt}{dy} = -(y + 1)$

$\frac{dt}{dy} = -y - 1$

$t = \int{-y - 1\,dy}$

$t = -\frac{1}{2}y^2 - y + C$.

You also know $y(0) = 2$...

$0 = -\frac{1}{2}(2)^2 - 2 + C$

$0 = -2 - 2 + C$

$0 = -4 + C$

$C = 4$.

So the solution is

$t = -\frac{1}{2}y^2 - y + 4$

$-2t = y^2 + 2y - 8$

$-2t = y^2 + 2y + 1^2 - 1^2 - 8$

$-2t = (y + 1)^2 - 9$

$9 - 2t = (y + 1)^2$

$y + 1 = \pm \sqrt{9 - 2t}$

$y = -1 \pm \sqrt{9 - 2t}$.
• Feb 10th 2010, 01:26 AM
kpizle
Quote:

Originally Posted by Prove It
$y'(t) + \frac{1}{y(t) + 1} = 0$

$\frac{dy}{dt} = -\frac{1}{y + 1}$

$\frac{dt}{dy} = -(y + 1)$ RIGHT HERE, I MADE A WRONG TURN

$\frac{dt}{dy} = -y - 1$

$t = \int{-y - 1\,dy}$

$t = -\frac{1}{2}y^2 - y + C$.

You also know $y(0) = 2$...

$0 = -\frac{1}{2}(2)^2 - 2 + C$

$0 = -2 - 2 + C$

$0 = -4 + C$

$C = 4$.

So the solution is

$t = -\frac{1}{2}y^2 - y + 4$

$-2t = y^2 + 2y - 8$

$-2t = y^2 + 2y + 1^2 - 1^2 - 8$

$-2t = (y + 1)^2 - 9$

$9 - 2t = (y + 1)^2$

$y + 1 = \pm \sqrt{9 - 2t}$

$y = -1 \pm \sqrt{9 - 2t}$.

Oh my gosh!!

You flipped everything!!!! Of course!

I have been beating this thing for about two hours now and thought I had exhausted all the tricks.

My goodness, that simplies things.

Thank you so very much! I really appreciate it.
• Feb 10th 2010, 01:45 AM
Prove It
Quote:

Originally Posted by kpizle
Oh my gosh!!

You flipped everything!!!! Of course!

I have been beating this thing for about two hours now and thought I had exhausted all the tricks.

My goodness, that simplies things.

Thank you so very much! I really appreciate it.

Yes, I like the flipping method.

Another alternative is to separate the variables.

$\frac{dy}{dt} = -\frac{1}{y + 1}$

$(y + 1)\,\frac{dy}{dt} = -1$

$\int{(y + 1)\,\frac{dy}{dt}\,dt} = \int{-1\,dt}$

$\int{(y + 1)\,dy} = \int{-1\,dt}$.

You should get the same answer.