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Differential Equation: Check my work please

  1. Aug 13, 2010 #1
    I just want someone to check my work this is the problem

    (3x^2 − y)dx + (y − x)dy = 0

    I'm going to use the exact equation method to solve it because it seems like the most practical method for the case...I was thinking substitution but the power is not the same for the highest one.

    M=3x^2 − y
    N =y − x

    dM/dy= -1 = dN/dx

    fx = 3x^2 - y dx
    f = [tex]\int(3x^2 - y)dx[/tex] = x^3 - xy +h(y)
    fy =S (y-x)dy = y^2/2 -xy + g(x)
    h(y)= x^3+c
    g(x)= y^2/2+c

    Implicit solution is

    x^3-xy+y^2/2 +c = 0 or xy-x^3-y^2/2 = c
     
  2. jcsd
  3. Aug 13, 2010 #2

    vela

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    Looks good. You can always check your answer by differentiating it and seeing if you get what you started with.
     
  4. Aug 13, 2010 #3
    Thanks, now I'm trying to figure out a new problem (didn't wanna make another post)
    y'+y/2=x/2y.

    I'm trying to figure out what would be a good method to solve this. Any suggestions? I ruled out exact. It looks like a linear equation kinda...but I'm not sure
     
  5. Aug 13, 2010 #4

    vela

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    Hint: Consider (y2)'.
     
  6. Aug 13, 2010 #5
    wel i know i could have dy/dx = x-y^2/2y But I still don't know which method to solve this. if i try substitution it doesn't work...

    Ok I tried this I substituted

    u=y^2
    du=2ydy

    2ydy=(x-y^2)dx
    du=(x-u)dx
    du/dx+u=x
    Makes a Linear Eq
    p(x)=1
    f(x)=x

    e^integral(1)dx = e^x

    e^x(u)=integral(e^x(x))dx
    e^x(u)=e^x+e^x(x)+c
    u=1+x+ce^-x
    y^2=1+x+ce^-x
    y=sqrt(1+x+ce^-x)

    :D right?
     
    Last edited: Aug 13, 2010
  7. Aug 13, 2010 #6

    vela

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    Almost. It doesn't quite work when you substitute it back in the original equation.
     
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