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Solve this DE using homogeneous equations

  1. Jun 13, 2012 #1
    1. The problem statement, all variables and given/known data
    dy/dx = (6x^(2)+xy+6y^(2))/(x^2)


    2. Relevant equations
    v = y/x
    y' = v + xv'


    3. The attempt at a solution

    y' = tan(6ln(abs(x))-C)/x ===> apparently not correct
     
  2. jcsd
  3. Jun 13, 2012 #2

    Curious3141

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    Homework Helper

    Tough to see where you went wrong if you don't show all your working.

    In your final line, I presume you meant [itex]y[/itex] instead of [itex]y'[/itex] (the latter means derivative of [itex]y[/itex]).

    Your answer looks almost correct, except that in your last step in going from [itex]v[/itex] to [itex]y[/itex] you may have divided by [itex]x[/itex] instead of multiplying by [itex]x[/itex]. Remember that [itex]v = \frac{y}{x}[/itex], so [itex]y = vx[/itex].

    Another pointer (not that your post is wrong in this respect, but this makes things neater) is that when you have a ln term plus an arbitrary constant, it's neater to express it as [itex]\ln|kx|[/itex]. This is because [itex]\ln|kx| = \ln|x| + \ln|k|[/itex], which is equivalent in form to [itex]\ln|x| + C[/itex].

    So your final answer should be [itex]y = x\tan(6\ln|kx|)[/itex], where k is an arbitrary constant.

    PS: also, to correct a possible misconception in your thread title, you're not solving the DE "using" homogeneous equations. The DE itself is a homogeneous equation, the definition of which is that multiplying all variables (both [itex]x[/itex] and [itex]y[/itex] in this case) by a constant scaling factor will not change the equation.
     
    Last edited: Jun 13, 2012
  4. Jun 14, 2012 #3

    HallsofIvy

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    You are correct up to the very last step. y= xv, not v/x!
     
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