Results 1 to 9 of 9

Math Help - Homogeneous or not.

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    14

    Homogeneous or not.

    To solve non homogeneous differential equations we first have to convert it to homogeneous differential equations (some cases).


    So I encountered its problems, or the examples of this type given in my source.


    After replacing x + y as u we got the equation -


    (2u + 1)du = (3u + 2)dx


    In another example (after assuming x + 2y as u)-
    (2u + 3)du = (4u + 5)dx


    None of the above 2 equations are homogeneous differential equations as claimed in the type's heading, do we have an explanation for this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by dE_logics View Post
    To solve non homogeneous differential equations we first have to convert it to homogeneous differential equations (some cases).


    So I encountered its problems, or the examples of this type given in my source.


    After replacing x + y as u we got the equation -


    (2u + 1)du = (3u + 2)dx


    In another example (after assuming x + 2y as u)-
    (2u + 3)du = (4u + 5)dx


    None of the above 2 equations are homogeneous differential equations as claimed in the type's heading, do we have an explanation for this?
    What were the two original equations?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    14
    Equation 1 -

    <br />
\frac{dy}{dx} = \frac{x + y + z}{2x + 2y + 1}<br />

    Equation 2 -
    <br />
\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}<br />
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member roshanhero's Avatar
    Joined
    Aug 2008
    Posts
    180
    In general I think eqn 2 will be the best one to explain this.
    You can change it into homogeneous simply by substituting x=X+h and y=Y+k.
    So after differeniating both you would get dx=dX and dy=dY
    After substituting it into the eqn you would get
    dY/dX=(X+h)+2(Y+k)+1/2(X+h)+4(Y+K)+3
    Separate all the X,Y,h and k terms.
    After that I hope you would be able to proceed
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    14
    aaa...well...actually, these are the procedures followed when
    <br />
\frac{a}{a1} <> \frac{b}{b1}.<br />
    And you gave the solution, I was asking why (2u + 3)du = (4u + 5)dx is considered homogeneous in my source, I mean they converted
    <br /> <br />
\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}

    " alt="

    " />

    to
    (2u + 3)du = (4u + 5)dx to make the question homogeneous, which it clearly is not.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2008
    Posts
    14
    Now I got no idea why <font face="monospace"><em>[
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    14
    C'mon people, pls!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by dE_logics View Post
    Equation 2 -
    <br />
\frac{dy}{dx} = \frac{x + 2y + 1}{2x + 4y + 3}<br />
    Let u=x+2y\implies\frac{\,du}{\,dx}=1+2\frac{\,dy}{\,d  x}\implies \frac{\,dy}{\,dx}=\tfrac{1}{2}\left(\frac{\,du}{\,  dx}-1\right)

    Thus, the DE becomes:

    \tfrac{1}{2}\left(\frac{\,du}{\,dx}-1\right)=\frac{u+1}{2u+3}

    Solving for \frac{\,du}{\,dx}, we get \frac{\,du}{\,dx}=\frac{2u+2}{2u+3}+\frac{2u+3}{2u  +3}\implies \frac{\,du}{\,dx}=\frac{4u+5}{2u+3} \implies (2u+3)\,du=(4u+5)\,dx\implies (4u+5)\,dx-(2u+3)\,du=0

    From looking at it, its not homogeneous.

    To test and see if a DE of this form is homogeneous do this:

    Since our DE is in the form of M(x,u)\,dx+N(x,u)\,du=0

    Evaluate M(tx,tu) and see if you end up with t^{\alpha}M(x,u) and evaluate N(tx,tu) and see if you end up with t^{\alpha}N(x,u). If you can get the powers to be the same, then its homogeneous. If you don't get the same exponents, then its not homogeneous. Its also non-Homogeneous when you can't get M(tx,tu)=t^{\alpha}M(x,u) or N(tx,tu)=t^{\alpha}N(x,u)


    However, if you need to solve this, you would use separation of variables. Even if you were to solve this, I don't think you will end up with an explicit solution.

    --Chris
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2008
    Posts
    14
    Actually I'm having problems in that technique too

    Since a^alpha will yield all numbers (accept something opposite to the polarity of a).

    So according to this, almost all equations are homogeneous!



    This problem was already listed in my long doubt folder.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Homogeneous ODE
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: December 3rd 2011, 08:47 AM
  2. Need help with homogeneous dif. eq.
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 2nd 2011, 04:14 AM
  3. Replies: 1
    Last Post: December 15th 2010, 05:43 AM
  4. Non homogeneous PDE
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: April 14th 2010, 06:34 AM
  5. Homogeneous de
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2008, 05:08 AM

Search Tags


/mathhelpforum @mathhelpforum