1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A simple DE.

Tags:
  1. Jun 4, 2017 #1
    1. The problem statement, all variables and given/known data

    ##(2x + 3y + 1)dx + (4x + 6y + 1) dy = 0##

    ##y(-2) = 2##

    2. Relevant equations


    3. The attempt at a solution

    Let ##z = 2x + 3y##
    then ##z^\prime = 2 + 3y^prime##

    ##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0##
    ##\dfrac{dz}{dx} = \dfrac{z- 1}{2z + 1}##

    ##x = 2z + 3\log(1 - z) + C = 2(2x + 3y) + 3\log|(1 - 2x - 3y)| + C##

    Plugging the values, I get ##C = -6##

    So, ##0 = x + 2y + log|3y + 2x - 1| - 2##

    But the answer given is ##0 = x + 2y \color{red}{-} log|3y + 2x - 1| - 2##.

    Is the answer in the book wrong ?
     
  2. jcsd
  3. Jun 4, 2017 #2

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    @Buffu I looked over your solution=I don't see any mistakes.
     
  4. Jun 4, 2017 #3

    Buzz Bloom

    User Avatar
    Gold Member

    Hi Buffu:

    I think your equation where dz/dx = f(x)is wrong.
    You want to do two things first.
    1. Substitute z=2x+3y into the given equation.
    2. Substitute an expression for dy=g(dz) into the result of step 1.

    Hope that helps.

    Regards,
    Buzz
     
  5. Jun 5, 2017 #4
    ##\displaystyle \dfrac{(z + 1)}{2z + 1} + \dfrac13\left({dz \over dx} - 2\right) = 0 \iff \dfrac{(3z + 3) - 4z - 2}{2z + 1} + {dz \over dx} = 0\iff \dfrac{ 1 - z}{2z + 1} + {dz \over dx} = 0 ##
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted