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Solving a DE with linear coefficients

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex](2x-y)dx+(4x+y-6)dy=0[/tex] I need to solve by recognizing that I have linear coefficients


    2. Relevant equations
    [tex]2x-y=0[/tex]
    [tex]4x+y-6=0[/tex]
    Solving this sytem of equations I get x=1 and y=2


    3. The attempt at a solution
    After finding my intersection point to be (1,2) I let u=x-1, v=y-2. Which gives, x=u+1 dx=du, and y=v+2 dy=dv
    I substitute back for x and y and get [tex](2(u+1)-(v-2))du+(4(u+1)+(v+2)-6)dv=0[/tex] I simplified and got [tex](2u-v)du+(4u+v)dv=0[/tex] Now I have a homogeneous function of degree 1. so I let [tex]w={\frac{u}{v}}[/tex] which gives u=vw du=vdw+wdv
    making that substitution and dividing by v gives [tex](2w-1)(vdw+wdv)+(4w+1)dv=0[/tex] Which gives [tex](2w^2+3w+1)dv+v(2w-1)dw=0[/tex]
    This is where i get stuck. Now, i seperate the variables. [tex]∫{\frac{1}{v}}dv+∫{\frac{2w-1}{2w^2+3w+1}}dw=0[/tex] The answer in the back of the book is x+2y+c=3ln(x+y+2), the book is notorious for over simplifying their final answer but, even then. I can't get there. I've tried partial fraction decomposition, observation, everything I can think of, maybe my brain is just tired and I'm missing something really obvious.

    EDIT: I was right on one account. Im really tired and i missed something obvious. I was looking at the wrong answer in the back. The actual answer is [tex](x+y-3)^3=c(2x+y-4)^2[/tex] I'll go back and see if I can't get this answer after I finish the problem I'm on.

    Update: I did get retry this and I got [tex](y+2x-4)^3=c(y+x-3)^2[/tex] My answer and the books answer are different in that the exponents are swapped. Now I can't see how they got their exponents opposite of mine, and its late. So I'm gonna chalk this up as a typo and call it a win.

    I solved this by doing partial fractions, but instead of using w=u/v I used w=v/u. This made it cleaner. It took way too much algebra to get from the raw anti-derivative to the answer I got.

    Just becuase I'm kinda proud of this, and if someone see's an algebra mistake that would yield what the book has, then all the better.

    After doing partial fractions and substiuting back for x and y I get:
    [tex]ln(x-1)+3ln({\frac{y-2}{x-1}}+2)=2ln({\frac{y-2}{x-1}}+1)+lnc[/tex]
    Combining my ln's i get [tex]ln(({\frac{y-2}{x-1}}+2)^3(x-1))=ln(c({\frac{y-2}{x-1}}+1)^2)[/tex]Writing both sides as a power of e I get: [tex]({\frac{y-2}{x-1}}+2)^3(x-1)=c({\frac{y-2}{x-1}}+1)^2[/tex]Writing both sides with a common denominator I get:[tex]{\frac{(y+x-4)^3}{(x-1)^3}}(x-1)=c{\frac{(y+2x-3)^2}{(x-1)^2}}[/tex] Now I multiply through by (x-1)2 And I get my final answer [tex](y+x-4)^3=c(y+2x-3)^2[/tex]

    Man, that was fun! :)

    Another Update: When copying the problem down on my homework, I noticed an error in my partial fractions. Where I had my values for A and B switched. So by fixing that mistake, I did end up with the exact answer in the back of the book! Hooray!
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2
    Condensed to one post.
     
    Last edited: Mar 6, 2013
  4. Mar 6, 2013 #3
    Combined with OP.
     
    Last edited: Mar 6, 2013
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