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Homework Help: Acceptible answer?

  1. Aug 18, 2008 #1
    starting with question of find the general sollution of the differential equation

    X2y'=y2+3xy+X2 would an acceptable answer be y=-x(ln|x|+c+1) i would show all my working but my camera isnt working so i'll save you must the trouble and just skip to a part that i know is correct where dx/x=(v+1)-2dv

    where v=y/x. thanks in advance i think its right just want a second oppinion before plowing it into an exam
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 18, 2008 #2


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    Homework Helper

    I don't understand your notation. Is 'X' different from 'x'?
  4. Aug 19, 2008 #3
    no just a bit daft with the caps lock button
  5. Aug 19, 2008 #4


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    Well, if y= y=-x(ln|x|+c+1), then y'= -(ln|x|+ c+1)- 1. Putting those into the differential equation, x2y'=y2+3xy+x2 would give you x2(-ln|x|+ c+ 1)- 1)= (ln|x|+ c+ 1)2+ 3x(-ln|x|+ c+ 1)+ x2. Since the right side is clearly going to involve "(ln|x|)2", I don't see how those are going to be equal.

    I suspect you have integrated incorrectly. Yes, this is a homogenous equation and the substitution v= y/x gives
    [tex]\frac{dv}{(v+1)^2}= \frac{dx}{x}[/tex]
    Integrating both sides of that gives
    [tex]-\frac{1}{v+1}= ln|x|+ C[/tex]
    [tex]v+1= -\frac{1}{ln|x|+ C}[/tex]
    [tex]\frac{y}{x}= -1-\frac{1}{ln|x|+ C}[/tex]
    [tex]y= -x-\frac{x}{ln|x|+ C}[/tex]
  6. Aug 19, 2008 #5


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    I worked it out and did not get the same answer as you did. ln(x)+c is supposed to be the denominator of some fraction in y.
  7. Aug 19, 2008 #6
    got it now, was just checking more to see if it would be acceptible to leave logs and things in the answer but made an error in the question
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