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Homework Help: Dif. eq. problem - dont know why it is wrong

  1. Feb 4, 2006 #1
    dif. eq. problem -- dont know why it is wrong

    Solve for y: [tex]x \frac{dy}{dx} + 6y=5x[/tex]

    with y(1)=2

    [tex]\frac{dy}{dx} =5 - 6\frac{y}{x} [/tex]
    [tex]\frac{6y}{x} \frac{dy}{dx} =5[/tex]
    [tex] 3y^2=\frac{5}{2}x^2 +C[/tex]
    [tex] y= \sqrt{\frac{5x^2}{6}+\frac{C}{3}}[/tex]

    solve for when y(1)=2



    [tex]y= \sqrt{\frac{5x^2}{6}+\frac{19}{6}}[/tex]

    what did i do wrong? because this is not the answer
    Last edited: Feb 4, 2006
  2. jcsd
  3. Feb 4, 2006 #2
    y/x does not equal (dy)/(dx), You need to divide by x and then find an integrating factor.
  4. Feb 4, 2006 #3
    sorry, that was typo! could you check it now?
  5. Feb 4, 2006 #4
    It looks even worse, you somehow make addition into multiplaication. I already told you what you need to do. You need to divide everything by x to get the y' by itself and then find an integrating factor, you should know how to do that if you're being given this kind of problem because it certainly isn't separable,
  6. Feb 4, 2006 #5
    oh thanks. by the way, when I am given a dif. eq. and they ask me "What are the constant solutions of this equation? "

    what exactly are do they want me to find?
  7. Feb 4, 2006 #6


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    As said before, you're taking the wrong approach.

    You need to divide it by x to get it into standard form or whatever it's called, and then get an integrating factor.

    I get an answer of y = 5x/7 +c/x6
    Last edited: Feb 4, 2006
  8. Feb 4, 2006 #7
    Umm... I think they might mean for you to find the solutions that are just a constant function ie. y=c c is just some constant.
  9. Feb 4, 2006 #8
    You forgot to divide c by x6
  10. Feb 4, 2006 #9


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    Ya. Actually I forgot to put in period and added it in haphazardly without thinking. I'm quite absent minded some times. I'll edit my post. I edited my last post.
    Last edited: Feb 4, 2006
  11. Feb 4, 2006 #10

    [tex] x\frac{dy}{dx}+6y=5x[/tex]

    [tex] \frac{dy}{dx}+\frac{6}{x}y=\frac{5}[/tex]

    then your integrating factor is


    multiply through by integrating factor

    [tex] x^6\frac{dy}{dx}+6x^5y=5x^6[/tex]

    then integrate both sides

    [tex]x^6y=\frac{5x^7}{7}+C [/tex]

    Divide through by x^6


    do you see where your mistake is?

    now solve for C

    [tex] y(1)=2 [/tex]

    [tex] 2=\frac{5}{7}1+C [/tex]

    [tex] 2-\frac{5}{7}=C [/tex]

    [tex] \frac{9}{7}=C [/tex]

    so the specific solution is

    Last edited: Feb 4, 2006
  12. Feb 4, 2006 #11
    The term on the right hand side should be 5x not 5 so your answer is wrong as well.
  13. Feb 4, 2006 #12
    shizer your right

    Last edited: Feb 4, 2006
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