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Differential Eq. (Substitution)

  1. Feb 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Solve by making an appropriate substitution. I am given the homogeneous DE:


    [tex]xdx+(y-2x)dy=0[/tex]

    Now we have bee using either y=ux or x=vy. . . I tried both, but the latter seemed easier.

    [tex]x\frac{dx}{dy}+y-2x=0[/tex] letting x=vy and dx/dy=v+y*dy/dv

    [tex]vy(v+y\frac{dy}{dv})+y-2vy=0[/tex]

    [tex]v^2+y^2\frac{dy}{dv}+y-2vy=0[/tex]

    Here is where I get stumped. . . this is supposed to be separable now right? Because I can't seem to see it.

    A hint would be swell!
     
    Last edited: Feb 20, 2008
  2. jcsd
  3. Feb 21, 2008 #2
    I think you need to check that again.
     
  4. Feb 21, 2008 #3
    [tex]x\frac{dx}{dy}+y-2x=0[/tex], lets divide by x to get

    [tex]\frac{dx}{dy}+\frac{y}{x}=2[/tex] or

    [tex]\frac{dx}{dy}+(\frac{x}{y})^{-1}=2[/tex] now lets take the sub

    [tex] v= \frac{x}{y}, x=vy, \frac{dx}{dy}=v+y\frac{dv}{dy}[/tex],(note: you made a mistake here) now let's go back and substitute we get

    [tex]v+y\frac{dv}{dy}+\frac{1}{v}=2[/tex] so we get

    [tex]y\frac{dv}{dy}+\frac{v^{2}+1}{v}=2[/tex]

    [tex]y\frac{dv}{dy}=\frac{2v-v^{2}-1}{v}=-\frac{(v-1)^{2}}{v}[/tex]

    Now this is separable and i think you will be fine from here on, right?
     
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