1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.


    [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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook