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Differential equation - seperation of variable

  • Thread starter kylerawn
  • Start date
  • #1
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Homework Statement



e^3x dy/dx = 1/2y

Homework Equations





The Attempt at a Solution



e^3x dy/dx = 1/2y

e^3x/dx = 1/2ydy

I can't determine the dirivitives for this equation can someone help me :)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
116
1
Try rearranging the equation so that dx and dy are in the numerator instead of the denominator.
 
  • #3
8
0
I tried that but i have the correct answer if y(0) = 4 the answer to the problem is

sqrt(16.33 - .33e^-3x)

if i rearange so dx and dy are in the numerator I get

dx/e^3x = dy/2y
 
  • #4
Tom Mattson
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dx/e^3x = dy/2y
Nope, you're being careless with the algebra.
 
  • #5
116
1
dx/e^3x = dy/2y
I am getting a different equation. Try rearranging again. :)
 
  • #6
8
0
dx/e^3x = 2ydy

1/e^3x = y^2 + C

y= sqrt(C+e^-3x)

4= sqrt (C+ e^-3(0))

16 = C+1

C = 15

Thats what I keep getting I know im messing up somehwere
 
  • #7
116
1
Double check your integration of 1/e^(3x).
 
  • #8
Tom Mattson
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And try writing it as:

[tex]\exp^{-3x}dx[/tex]

before you integrate.
 
  • #9
8
0
And try writing it as:

[tex]\exp^{-3x}dx[/tex]

before you integrate.
Thanks alot

[tex]\exp^{-3x}dx[/tex]

[tex]\-.33exp^{-3x}[/tex]
 
  • #10
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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Not quite, you're missing a negative sign. And instead of 0.33, you really should have 1/3. How would a rounded decimal emerge anyway? If you did the u-substitution that is required, the -1/3 should pop right out.
 
  • #11
8
0
right :) thanks forgot the negitive in front. I realize it should be -1/3 i was just going with the answer my prof gave us :)

Thanks for all your help
 
  • #12
8
0
I came across another problem I cant solve

[tex]dy/dx=5y+1[/tex]

[tex]dy/5y+1=dx[/tex]

[tex]exp^{5y+1} = x[/tex]

but i know this is incorrect can you point out where i am going wrong?
 

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