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Help on differential equations PLEASE

  1. Sep 15, 2008 #1
    1. I need to reduce [tex](a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}x +c_{2})dy = 0[/tex] to:

    [tex]\frac{dy}{dx} = F ( ax + by) [/tex]

    with [tex] a_{1}b_{2} = a_{2}b_{1}[/tex]

    i.e.:
    [tex] a_{1}b_{2}=a_{2}b_{1}[/tex]

    [tex]\frac{a_{2}}{a_{1}} = k[/tex] and [tex]\frac{b_{2}}{b_{1}} = k[/tex]

    2.
    First Try: I solved for dy/dx and tired to deal with the a's and b's. But I am still left with c1 and c2.
    Second Try: I tried a different approach by letting x=m+h and y=n+k, then substitute it in and try to solve for h and k. But since the coefficient of dy has 2 x's instead of a x and y, i got a system of 2 "ugly" equations that lead nowhere... (or does it?)

    I tried to used the 2 methods in different ways but it seems like I am running in circle.. Can anyone pls help??
     
  2. jcsd
  3. Sep 15, 2008 #2

    gabbagabbahey

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    Here's a hint: Would you consider [tex]\frac{u+2}{5u+10}[/tex] to be a function of [tex]u[/tex]?

    How about [tex]\frac{u+c_1}{ku+c_2}[/tex] where [tex]c_1,&c_2,&k[/tex] are constants?

    What about when [tex]u \equiv ax+by[/tex] where [tex]a \equiv a_1[/tex] and [tex]b \equiv b_1[/tex]?
     
  4. Sep 15, 2008 #3

    tiny-tim

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    Hi chocok! :smile:

    It must mean [tex](a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}y +c_{2})dy = 0[/tex]

    what you've been given is obviously a misprint! :redface:

    And do you need to get rid of the constants, c1 and c2? …

    for example, (ax + by + c1)2 + sin(ax + by + c2) + log(ax + by + c3) would still be of the form F(ax + by). :smile:
     
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