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A simple DE.

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  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

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