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Solving Bernoulli's Differential Equation

  1. Oct 5, 2016 #1
    1. The problem statement, all variables and given/known data
    xy(dx)=(y2+x)dy


    2. Relevant equations
    integrating factor : u(x)=e∫p(x)dx
    standard form of linear DE: dy/dx + P(x)y=Q(x)
    standard form of bernoulli's differential equation: dy/dx + P(x)y=Q(x)yn
    change of variables v=y1-n
    3. The attempt at a solution
    xy(dy)=(y2+x)dx
    xy(dy/dx)=y2 +x
    dy/dx=y/x +1/y
    dy/dx-y/x=y-1

    yn=y-1
    y-n=y1

    multipliying both sides by y1 yeilds:
    y1(dy/dx) -y2/x=1

    using change of variable v=y1-n=y2
    dv/dx=(2y)(dy/dx)
    (1/2)(dv/dx)=(y)(dy/dx)
    and subbing the results yeilds:
    (1/2)(dv/dx)-v/x=1

    using an integrating factor to solve the resulting linear differential equation
    u(x)=e∫p(x)dx
    u(x)=e-∫(1/x)dx=1/x

    multiplying both sides of the equation by the integrating factor yeilds:
    (1/x)[(1/2)(dv/dx)-v/x]=1/x
    (d/dx)[v/x]=1/x

    integrating both sides:
    (1/2)∫(v/x)d=∫1/xdx
    (1/2)v/x=ln|x| + C
    v=2ln|x| +C

    now subbing back in y2]
    y2]=2ln|x| + C

    Unfortunately the answer is :

    y2=-2x+Cx2

    where did I go wrong?
    Thanks ahead of time, you guys rock
     
  2. jcsd
  3. Oct 5, 2016 #2

    LCKurtz

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    Gold Member

    Your equation is not in the proper form for the integrating factor. The leading coefficient must be ##1##, so you need to multiply your equation by ##2## before continuing.
     
  4. Oct 6, 2016 #3
    thanks a bunch, I got it now
     
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