# Thread: differential equations first order

1. ## differential equations first order

dy/dx = xy^2/x+5

how do I algebraically convert this into the form y/x

2. ## Re: differential equations first order

Originally Posted by Dfaulk044
dy/dx = xy^2/x+5

how do I algebraically convert this into the form y/x
I'm not really sure what you are asking here. You aren't going to be able to express this as f(y/x).

The method you would use to solve this is called separation of variables.

$\displaystyle \frac{\text{dy}}{\text{dx}}=\frac{x y^2}{x+5}$

$\displaystyle \frac{\text{dy}}{y^2}=\frac{\text{dx} x}{x+5}$

And you integrate on both sides to obtain your solution. You should be able to do this. Post back if you need more help.

3. ## Re: differential equations first order

is the answer -y^-1 = ln[x+5]

4. ## Re: differential equations first order

Originally Posted by Dfaulk044
is the answer -y^-1 = ln[x+5]
It's not hard to check it. The derivative of the left side, with respect to x, is y^{-2} dy/dx and the derivative of the right side is 1/(x+ 5).
So y^{-2} dy/dx= 1/(x+ 5) so dy/dx= y^2/(x+ 5).

I do NOT see an "x" in the numerator on the right.

The integral of $\displaystyle x/(x+ 5)$ is NOT ln(x+ 5). If you need to, make the substitution u= x+ 5 so that x= u- 5 and $\displaystyle \frac{x}{x+ 5}= \frac{u- 5}{u}= 1- \frac{1}{u}$.