1. ## differential help

solve when x=0 and y= 2

ok I have setup as shown(is this ok?). is there some rule about F/F'? to make it easier to solve?

many thanks

2. ## Re: differential help

Originally Posted by decoy808
solve when x=0 and y= 2

ok I have setup as shown(is this ok?). is there some rule about F/F'? to make it easier to solve?

many thanks
$\frac{2y+4}{y^2+4y}dy=dx$

Integrating both sides,

$ln(y^2+4y)=x+C$

Substitute x=0 and y=2 to find C.

3. ## Re: differential help

\displaystyle \begin{align*} (2y + 4)\,\frac{dy}{dx} &= y^2 + 4y \\ \left[\frac{2y + 4}{y(y + 4)}\right]\,\frac{dy}{dx} &= 1 \\ \int{\left[\frac{2y + 4}{y(y+4)}\right]\,\frac{dy}{dx}\,dx} &= \int{1\,dx} \\ \int{\frac{2y + 4}{y(y + 4)}\,dy} &= \int{1\,dx}\end{align*}

Go from here, you will need to use a substitution or Partial Fractions on the LHS.

4. ## Re: differential help

Originally Posted by decoy808
solve when x=0 and y= 2

ok I have setup as shown`(is this ok?). is there some rule about F/F'? to make it easier to solve?

many thanks
No, there is no rule about F/F'. There is, however, a rule about F'/F which this is. If you let u= F, then du= F' dx so
$\int \frac{F'}{F}dx= \int \frac{1}{u}du= ln|u|+ C= ln|F(x)|+ C$.