# Thread: First order diff eqns, the -ive and +ive signs?

1. ## First order diff eqns, the -ive and +ive signs?

Hey guys im stuck as to the origins of the + and minus signs of the following question? (It is attached via a jpeg)

The answer tells me that u

int/ dy/y(y+1) = int/ dx/ x(x-1)

int/ (1/y - 1/y+1) dy = int/ (-1/x + 1/x-1)

ok i dont get this part, where do they get the -ive sign for the 1/x and how do they decide that it is -ive plus sign for the int of the y side and +ive sign for the x side of the eqn. Am i missing a really obvious point? Please help me, im fine with the eqns but the signs are giving completely diff answers

2. What you are doing here is simply splitting the equation into its y part and its x part. So you have
$\frac{dy}{dx}=\frac{y(y+1)}{x(x-1)}$

[\math]\frac{1}{y(y+1)}dy=\frac{1}{x(x+1)}dx[/tex]

Integrate both sides. Fiddle about to get y. And then use y(2)=whatever to solve for the constant.

3. Originally Posted by Latkan

int/ dy/y(y+1) = int/ dx/ x(x-1)

int/ (1/y - 1/y+1) dy = int/ (-1/x + 1/x-1)
To get from here

$\int \frac{dy}{y(y+1)} = \int\frac{ dx}{ x(x-1)}$

to here

$\int \frac{1}{y} - \frac{1}{y+1}~ dy = \int \frac{-1}{x} + \frac{1}{x-1}~dx$

The denominators have been broken up using partial fractions. Do you know this method?

This might be worth a read

Partial fraction - Wikipedia, the free encyclopedia