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Solving a DE

  1. Sep 22, 2011 #1


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    1. The problem statement, all variables and given/known data
    I must solve [itex]x^2y'+y^2=xyy'[/itex].

    2. Relevant equations
    Not sure but I think that [itex]f(tx,ty)=t^nf(x,y)[/itex] could help.

    3. The attempt at a solution
    My first reflex was to define a new variable [itex]z=y^2[/itex] but I was stuck a few steps further.
    So I checked out if it was homogeneous and I found out that yes it is, of order 2.
    So I called a new variable [itex]v=y/x[/itex].
    After some algebra, I reached [itex]v(x)=Ae^{\int \frac{x^2-1}{x}dx}[/itex] where A is a constant. Now to get y(x), I'd multiply v(x) by x.
    I'm not asking if my result is correct (I might have made some errors but overall I think the method does work. I'll check the result tomorrow since it's already too late), rather if there's a nicer or faster way to solve the exercise.
    What would you have done in order to solve the DE?
  2. jcsd
  3. Sep 22, 2011 #2
    I have absolutely no idea how you found it was exact and I have to leave soon.

    My advice is to divide both sides by x^2 and see what you recognize. I have to sleep now my friend.

    *disappears from the shadows*
  4. Sep 23, 2011 #3
    It is a homogenous equation. Therefore, it should be alright if you substitute v=y/x
  5. Sep 23, 2011 #4


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    Well, both flyingpig and icystrike tell you this equation is homogenous and recomend the substitution v= y/x. But your post says that is exactly what you did!

    The only question is why you left
    [tex]\int \frac{x^2- 1}{x}dx= \int x dx- \int\frac{dx}{x}[/tex]
    as an integral rather than actually integrating it!
  6. Sep 23, 2011 #5


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    Thank you all guys, so my way was ok and it seems it's efficient.
    Well I said "homogeneous", not exact. I guess you were tired like me. :)
    I see. That's exactly what I've done so I'm happy it seems I'm grasping it.
    Yes... about the integral, I was so tired that when I saw the quotient I didn't even think 1 s about solving the integral. Now that I look at it, it looks like a piece of cake.
    Thanks once again guys, problem solved.
  7. Sep 23, 2011 #6


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    Hmm I'm turning insane.
    I reached [itex]v(x)=\frac{C}{x}e^{\frac{x^2}{2}}[/itex] which indeed satisfy the DE [itex]x^2(xv'+v)+v^2x^2-x^2v(xv'+v)=0[/itex] as it should.
    Now to get y(x), I thought I simply had to multiply v(x) by x, but this didn't work.
    If I do so, I reach [itex]y(x)=Ce^{\frac{x^2}{2}}[/itex]. This doesn't satisfy the original DE... I really don't understand why.
    The left hand side of the original DE becomes [itex]x^3Ce^{\frac{x^2}{2}}+C^2e^{x^2}[/itex]. The right hand side (which should be equal to the left hand side but isn't) is [itex]x^2C^2e^{x^2}[/itex].

    If you're interested in my y', it's [itex]y'(x)=xCe^{\frac{x^2}{2}}[/itex]. I have no idea what's going on!
  8. Oct 9, 2011 #7


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    Hey guys I still don't know what I did wrong. I posted my work in my last post and the answer doesn't satisfy the original DE...
    On the other hand I attacked the problem by a friend's approach.
    I divided the original DE by xy to get [itex]y' \left ( 1 - \frac{x}{y} \right ) -\frac{y}{x}=0[/itex]. Then [itex]y'=\frac{y}{x} \cdot \frac{1}{1-\frac{x}{y}}[/itex] which is separable. I get the implicit solution for y: [itex]\frac{y}{x}-\ln \left ( \frac{y}{x} \right )=\ln x+C[/itex].
    But I still haven't figured out my error(s) in my last post.
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