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Homework Help: Linear Equations DE

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm told that this is a linear equation

    [tex]y' = \frac{4ln|x| - 2x^2y}{x^3}[/tex]

    2. The attempt at a solution

    [tex]x^3 = (4ln|x|-2x^2y)\frac{dx}{dy}[/tex]

    giving:
    [tex]P = 4ln|x|-2x^2[/tex]
    [tex]Q = x^3[/tex]
    [tex]\int Pdx = 4x-\frac{2x^3}{3}[/tex]
    [tex]e^\int^p^d^x = e^4^x^-^\frac^{2x^3}^{3}[/tex]
    [tex]ye^\int^p^d^x = \int e^4^x^-^\frac^{2x^3}^{3} x^3[/tex]

    I know this is too complicated and ugly.. It has to be wrong so far...
     
    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 27, 2007 #2

    Dick

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    That is not a linear equation. Linear equations have the property that if y_1 and y_2 are solutions, then so is A*y_1+B*y_2. Your solution seems to consist of just throwing symbols around at random. The form of the equation doesn't make me think there is any way to get a closed form solution. If you really need to solve it you may have to try numerical techniques.
     
  4. Jan 27, 2007 #3
    my textbook describes it as "first order linear DE"

    I'm suppose to solve the equation by putting it into the form [tex]\frac{dy}{dx} + Py = Q[/tex]

    then integrate
     
  5. Jan 27, 2007 #4

    Dick

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    Ok. But then your assignments of P and Q don't look at all right. Concentrate of the form in 1) and try again.
     
  6. Jan 27, 2007 #5

    Dick

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    I see. It's an 'integrating factor' trick. Forgot.
     
  7. Jan 27, 2007 #6
    is that what it's called? because I can not find anything on the internet that helps me with I look for liner equation DE
     
  8. Jan 27, 2007 #7

    Dick

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    Just go by your book. But you've identified P and Q wrong.
     
  9. Jan 27, 2007 #8

    HallsofIvy

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    That is, in fact, a linear equation for y as a function of x. Rewriting to
    [tex]\frac{dx}{dy}[/tex] however gives you a non-linear equation for x as a function of y. There is no reason to do that.

    The equation
    [tex]\frac{dy}{dx}= \frac{4ln|x|-2x^2y}{x^3}[/tex]
    can be rewritten as
    [tex]\frac{dy}{dx}+ \frac{2}{x}y= \frac{4 ln|x|}{x^3}[/tex]

    Now you are looking for an "integrating factor" [itex]\mu(x)[/itex] such that
    [tex]\mu(x)\frac{dy}{dx}+ \mu(x)\frac{2}{x}y= \frac{d(\mu(x)y}{dx}[/tex]
    doing the derivative on the right, this reduces to
    [tex] \mu(x)\frac{2}{x}y= \frac{d\mu(x)}{dx}y[/tex]

    In other words we must have
    [tex]\frac{d\mu(x)}{dx}= \frac{2}{x}[/itex]
    That's an easy separable equation.

    Solve it to find [itex]\mu(x)[/itex] and the rest is easy.
     
  10. Jan 27, 2007 #9

    Dick

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    Mmm. I was waiting for him to identify the P and Q. Sorry about my confusing gaffe in saying it was not linear.
     
  11. Jan 28, 2007 #10

    HallsofIvy

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    S'awright!
     
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