# First Order Differential Equation

• Feb 18th 2009, 08:57 AM
Simplicity
First Order Differential Equation
Question:
$\displaystyle \frac{dy}{dt} = \frac{t+y}{t+y+1}$ Solve for $\displaystyle y(t)=...$

Approach:
Letting $\displaystyle u=t+y+1$ then getting $\displaystyle \frac{dy}{dt} = \frac{du}{dt}-1$ and substituting into main equation to get $\displaystyle \frac{du}{dt}-1 = \frac{u-1}{u}$ and rearraning to get $\displaystyle \frac{du}{dt} + \frac{1}{u} = 2$ however it's not in integration factor form. How would I solve it?
• Feb 18th 2009, 09:04 AM
Jester
Quote:

Originally Posted by Simplicity
Question:
$\displaystyle \frac{dy}{dt} = \frac{t+y}{t+y+1}$ Solve for $\displaystyle y(t)=...$

Approach:
Letting $\displaystyle u=t+y+1$ then getting $\displaystyle \frac{dy}{dt} = \frac{du}{dt}-1$ and substituting into main equation to get $\displaystyle \frac{du}{dt}-1 = \frac{u-1}{u}$ and rearraning to get $\displaystyle \frac{du}{dt} + \frac{1}{u} = 2$ however it's not in integration factor form. How would I solve it?

Try
$\displaystyle \frac{du}{dt} = 2 - \frac{1}{u} = \frac{2u-1}{u}$

It's separable.
• Feb 18th 2009, 09:06 AM
Simplicity
Quote:

Originally Posted by danny arrigo
Try
$\displaystyle \frac{du}{dt} = 2 - \frac{1}{u} = \frac{2u-1}{u}$

It's separable.

As by:

$\displaystyle \int \frac{u}{2u-1} du = \int dt$?
• Feb 18th 2009, 09:07 AM
Mush
Quote:

Originally Posted by Simplicity
As by:

$\displaystyle \int \frac{u}{2u-1} du = \int dt$?

Yes.
• Feb 18th 2009, 09:10 AM
Simplicity
Quote:

Originally Posted by Mush
Yes.

What method am I using to solve $\displaystyle \int \frac{u}{2u-1} du$?
• Feb 18th 2009, 09:20 AM
Mush
Quote:

Originally Posted by Simplicity
What method am I using to solve $\displaystyle \int \frac{u}{2u-1} du$?

$\displaystyle p = 2u-1$

$\displaystyle dp/du = 2$

$\displaystyle \frac{1}{2}dp = du$

$\displaystyle \frac{1}{2} \int \frac{u}{p}dp$

$\displaystyle u = \frac{p+1}{2}$

$\displaystyle \frac{1}{2} \int \bigg(\frac{\frac{p+1}{2}}{p}\bigg)dp$

$\displaystyle \frac{1}{4} \int\bigg(\frac{p+1}{p}\bigg)dp$

$\displaystyle \frac{1}{4} \int \bigg(1+\frac{1}{p}\bigg) dp$