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Elementary Differential equations : seperable making it explicit

  • Thread starter Mdhiggenz
  • Start date
  • #1
327
1

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)
 

Answers and Replies

  • #2
3,812
92
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:
[tex]e^{\ln((x+4)/(y+2)^2)+y}=c[/tex]
which is equal to
[tex]\frac{x+4}{(y+2)^2} \cdot e^y=c[/tex]

And can you check the answer you posted? I get (-y+c) instead of (-y+1).
 
  • #3
327
1
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 ?
 
  • #4
3,812
92
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.
 
  • #5
327
1
Excellent, I appreciate your help on these problems.
 

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