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Differential Equations - Substitution

  1. Jun 8, 2009 #1
    Solve the Initial Value Problem:
    dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1

    I need to use substitution and use the formulas:
    du/dx = A + B(dy/dx)
    dy/dx = (1/B)*(du/dx - A)

    Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
    therefore A = 3, B = 2

    Pluging it into the formula gives me:
    (1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)

    Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.

    How do i make this equation a separable equation?
     
  2. jcsd
  3. Jun 8, 2009 #2
    (1/2)*(du/dx - 3) = (u)/(u + 2)
     
  4. Jun 8, 2009 #3
    1. The problem statement, all variables and given/known data
    Solve the Initial Value Problem:
    dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1


    2. Relevant equations
    I need to use substitution and use the formulas:
    du/dx = A + B(dy/dx)
    dy/dx = (1/B)*(du/dx - A)


    3. The attempt at a solution
    Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
    therefore A = 3, B = 2

    Pluging it into the formula gives me:
    (1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)

    Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.

    (1/2)*(du/dx - 3) = (u)/(u + 2)

    (u+2)/u du = 3/2 *2 dx

    integrate both sides to get:
    u + 2 ln |u| = 3x + c
    (3x + 2y) + 2 ln |(3x + 2y)| = 3x + c

    plugging in the initial conditions I got:
    -2 + 2 ln |5| = c

    therefore,
    (3x + 2y) + 2 ln |(3x + 2y)| = 3x - 2 + 2 ln |5|

    is this correct?
     
  5. Jun 8, 2009 #4

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi sami23! :smile:
    Noooo :redface:
     
  6. Jun 8, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is a question, not "learning materials". I am moving it.
     
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