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Homework Help: ODE by substitution

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data

    By means of substitution x=X+1, y=Y+2 ,shwo that the equation dy/dx=(2x-y)/(x+2y+5) can be reduced to dY/dX=(2X-Y)/(X+2Y).Hence, find the general solution of the given equation.

    2. Relevant equations



    3. The attempt at a solution

    The first part is quite simple to prove.

    Second part,

    since its a homogenous differential equation, i would use the substitution Y=vX here.

    v+X(dv/dX)=(2X-vX)/(2vX+X)

    [tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v+2} dv=\int \frac{1}{X} dx[/tex]

    ln |2v^2+2v+2|=-2ln |X|-c

    then v=Y/X

    ln |(2Y^2)/X^2+2Y/X+2|=-2ln |X|-c'

    Then substitute back again from the first part,

    the solution is

    ln |2(y-2)^2/(x-1)^2+2(y-2)/(x-1)+2|=-2ln |x-1|-c'

    AM i correct?
     
  2. jcsd
  3. Jul 1, 2010 #2
    Re: Ode

    check this again because i got

    [tex]-\frac{1}{2}\int \frac{4v+2}{2v^2+2v-2} dv=\int \frac{1}{X} dX[/tex]
     
  4. Jul 1, 2010 #3
    Re: Ode

    ok thanks ,how bout the rest?
     
  5. Jul 1, 2010 #4
    Re: Ode

    hmm, why do you change c to c' ?

    and, others correct if i'm not wrong.

    p/s: don't forget change the negative sign, :biggrin:
     
    Last edited: Jul 1, 2010
  6. Jul 1, 2010 #5
    Re: Ode

    oh, maybe its not necessary. I multiplied both sides with 2 so c is now 2c and c'=2c

    Is it ok to not change? Would it affect the equation?
     
  7. Jul 1, 2010 #6

    Mark44

    Staff: Mentor

    Re: Ode

    You did the right thing even though c and c' are arbitrary constants.
     
  8. Jul 1, 2010 #7
    Re: Ode

    yea, but you already multiply both side

    "ln |2v^2+2v+2|=-2ln |X|-c

    then v=Y/X

    ln |(2Y^2)/X^2+2Y/X+2|=-2ln |X|-c' "

    and you didn't multiplied here, correct me if i'm wrong
     
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