Elementary Differential equations : seperable making it explicit

[Mdhiggenz]

Homework Statement

(y+2)dx +y(x+4)dy=0

Homework Equations

The Attempt at a Solution

∫(1/x+4)dx + ∫(y/y+2)dy=0

ln(x+4)-2ln(y+2)+y=lnc

Here is where I get confused how do I make this into an explicit solution,

that +y really bothers me.

I was thinking ln [ (x+4)/(y+2)^2] +y = lnc

raising both sides to e^x

(x+4)/(y+2)^2+e^y=c

but the answer is

x+4 = [(y+2)^2]*e^(-y+1)

 

[Saitama]

I was thinking ln [ (x+4)/(y+2)^2] +y = lnc

raising both sides to e^x

(x+4)/(y+2)^2+e^y=c

That's not correct. When you raise both the sides to e, you get this:
$$e^{\ln((x+4)/(y+2)^2)+y}=c$$
which is equal to
$$\frac{x+4}{(y+2)^2} \cdot e^y=c$$

And can you check the answer you posted? I get (-y+c) instead of (-y+1).

 

[Mdhiggenz]

They hold that the answer is -y+c. I forgot to mention that it is a boundary problem y(-3)=-1

Would that have anything to do with it?

Also from what you got I do not quite understand how you got -y+c

if you multiply both sides by [(y+2)^2]/e^y

wouldnt it be x+4= c(y+2)^2/e^y ?

 

[Saitama]

wouldnt it be x+4= c(y+2)^2/e^y ?

Yep, it would be that, but 1/e^y=e^(-y).
c is a arbitrary constant, that is we can write it like this, e^c which is again a constant or we can even write it as ln(c) which is also equal to a constant. To get the answer you posted, i wrote c as e^c.

 

[Mdhiggenz]

Excellent, I appreciate your help on these problems.