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Solving a Differential Equation with Coefficients Linear in 2 variables.

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation:

    [tex](2x-y)dx+(4x+y-6)dy=0 [/tex]

    2. Relevant equations
    Solve for M and N as a linear system of equations; and

    [tex]x_t = u + h [/tex]
    [tex]y_t = v + k [/tex]

    3. The attempt at a solution

    M = 2x-y=0
    N = 4x+y-6=0

    2x=y
    4x+2x=6
    6x=6
    x=1
    y=2 ∴ [tex]x_t=u+1 \\ dx=du [/tex]
    [tex]y_t=v+2 \\ dy=dv [/tex]

    Substitute:

    (2(u+1)-(v+2))du+(4(u+1)+(v+2)-6)dv=0

    (2u-v)dw+(4u+v)dv=0 ;Homogeneous of degree one

    u=vw du=vdw+wdv

    (2vw-v)(vdw+wdv)+(4vw+v)dv=0
    Multiply out:

    [tex]\frac{(2w-1)dw}{(2w+1)(w+1)} + \frac{dv}{v} = 0 [/tex]

    Then partial fractions for the first term:

    [tex]\frac{(2w-1)}{(2w+1)(w+1)} = \frac{A}{2w+1} + \frac{B}{w+1}[/tex]

    Solved:

    A=3 B=-4

    Substitute A and B:

    [tex]\frac{3dw}{2w+1} + \frac{-4dw}{w+1}+ \frac{dv}{v} = 0 [/tex]

    Integrated:

    [tex]3ln|2w+1|-4ln|w+1|+ln|v|=C [/tex]

    My question: How exactly does the book get this answer: [tex](x+y-3)^3=c(2x+y-4)^2[/tex]


    Yes, I am bad at algebra.
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2

    sharks

    User Avatar
    Gold Member

    First, you must re-arrange your differential equation in this form:
    [tex]\frac{dy}{dx}=\frac{y-2x}{y+4x−6}=\frac{\frac{y}{x}-2}{\frac{y}{x}+4−\frac{6}{x}}[/tex]
    Since the equation above is not a function of [itex]\frac{y}{x}[/itex] only, it is an inhomogeneous equation. Hence, we seek a translation of axes: x=X+h and y=Y+k
    [tex]\frac{dY}{dX}=\frac{Y-2X+(k-2h)}{Y+4X+(k+4h-6)}[/tex]
    [tex]k-2h=0
    \\k+4h-6=0[/tex]
    Solving the above system for h and k give: h=1 and k=2
    So, we let x=X+1 and y=Y+2
    [tex]\frac{dY}{dX}=\frac{Y-2X}{Y+4X}=\frac{\frac{Y}{X}-2}{\frac{Y}{X}+4}[/tex]
    Since the above equation is now homogeneous, let [itex]\frac{Y}{X}=v[/itex]
    Then, [tex]\frac{dY}{dX}=v+X\frac{dv}{dX}[/tex]
    Solving, we get:
    [tex]\ln X= -2v + \ln (v+2)^2 +\ln (v+1)^3[/tex]
    [tex]2v=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
    [tex]2\frac{Y}{X}=\ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
    [tex]Y=\frac{X}{2} \ln \frac{(v+2)^2 (v+1)^3}{X}[/tex]
    Finally, replace v, Y and X with the original y and x.
     
    Last edited: Apr 21, 2012
  4. Apr 21, 2012 #3
    It's here that I am getting lost.

    I have also have [tex]\frac{dY}{dX}=v+X\frac{dv}{dX}[/tex] and set it equal to [tex]\frac{v-2}{v+4}[/tex] but when I separate the variables I get [tex]\frac{(v+4)dv}{v^2+3v+2}=\frac{-dX}{X}[/tex]

    Which does not integrate into something my textbook has.
     
  5. Apr 22, 2012 #4
    Never mind, I figured it out. It only took me a day. :P

    Also, thanks for everything. You were a huge help.
     
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