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Math Help - 1st order linear equation

  1. #1
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    1st order linear equation

    (x+y)dx=(-x)dy

    I know this equation requires a substitution of v=y/x at some point, but I cannot see how am supposed to use this. Can anyone help please .
    Last edited by Chris L T521; January 12th 2011 at 11:57 AM. Reason: Returned post to regular font size.
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  2. #2
    A Plied Mathematician
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    This is a homogeneous equation. That is, M and N are both homogeneous functions of the same degree. Hence, the substitution you mentioned will render the equation separable. You have y = vx. Then dy = v dx + x dv. Substitute in and simplify. What do you get?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    It is the standard substitution for homogeneus equations i.e. equations of the form

    P(x,y)dx+Q(x,y)dy=0

    where P and Q are homogeneus functions of degree d (in this case, d=1 ).

    Fernando Revilla

    Edited: Sorry, I didn't see Ackbeet's post.
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  4. #4
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    Quote Originally Posted by MathsLion View Post
    (x+y)dx=(-x)dy

    I know this equation requires a substitution of v=y/x at some point, but I cannot see how am supposed to use this. Can anyone help please .
    The idea behind that substitution is to transform the equation into another one that's separable. can you solve it now?
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  5. #5
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    Initial Condition

    Thank you very much.

    I solved the equation using this method, and obtained log(x)+C=y/x.

    I then checked this by using the attached file.

    However, the problem I am faced with is with that of the initial condition.

    If the initial condition is y(x=0)=-2, my constant keeps vanishing and do not know how to write my answer in EXPLICIT form.

    Thank you
    Attached Thumbnails Attached Thumbnails 1st order linear equation-maple-2-q5.pdf  
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  6. #6
    A Plied Mathematician
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    You're trying to apply the initial condition at a point at which the solution is invalid. Are you sure that's the correct initial condition?
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  7. #7
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    Quote Originally Posted by MathsLion View Post
    (x+y)dx=(-x)dy

    I know this equation requires a substitution of v=y/x at some point, but I cannot see how am supposed to use this. Can anyone help please .
    You're expected to learn something from your other threads: http://www.mathhelpforum.com/math-he...de-166809.html. If you did not understand what was told to you over there, you should have said so at that thread.
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