# Thread: First Order Differential Equation

1. ## 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?

2. 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.

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

It's separable.
As by:

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

4. Originally Posted by Simplicity
As by:

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

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

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

$\displaystyle dp/du = 2$

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

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

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

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

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

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