Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By HallsofIvy

Thread: First order ODE getting to homogenous with x and y substitutions

  1. #1
    Member
    Joined
    May 2009
    Posts
    243
    Thanks
    3

    First order ODE getting to homogenous with x and y substitutions

    Hi folks,


    by substituting $x = X + 1$ and $y = Y + 3 $ reduce the differential equation


    $ \dfrac{dy}{dx} = \dfrac{4x - y + 7}{2x + y - 1} $ ................... (1)


    to a homogeneous equation and hence find the general solution in terms of x and y.


    Substituting the values of x and y into (1) we get


    $ \dfrac{dy}{dx} = \dfrac{4X - Y}{2X + Y}$ which can be made homogeneous, that is, a function of $\frac{Y}{X} $ by dividing top and bottom by X.

    Once we have a homogeneous equation, it can be solved using the y = vx substitution, but first I have to get $ \frac{dy}{dx} $ into the form $ \frac{DY}{DX}$. I cannot see how to do this. I have tried different combinations of the chain rule, but not getting anywhere.

    A hint would really help!


    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,044
    Thanks
    3173

    Re: First order ODE getting to homogenous with x and y substitutions

    Your original equation is $\displaystyle \frac{dy}{dx}= \frac{4x- y+ 7}{2x+ y- 1}$. You make the substitution x= X- 1 (NOT "x= X+ 1"), y= Y+ 3. $\displaystyle \frac{dy}{dx}= \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}= \frac{dY}{dX}$. 4x-y+ 7= 4(X- 1)- Y- 3+ 7= 4X- Y and 2x+ y- 1= 2(X- 1)+ Y+ 3- 1= 2X+ Y. Yes, that makes the equation $\displaystyle \frac{dY}{dX}= \frac{4X- Y}{2X+ Y}$.

    As to how I went from $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dY}{dx}$, first I replaced y by Y+ 3: $\displaystyle \frac{dY}{dx}= \frac{Y+ 3}{dx}= \frac{dY}{dx}+ 0= \frac{dY}{dx}$. Now to go from $\displaystyle \frac{dY}{dx}$ to $\displaystyle \frac{dY}{dX}$, use the chain rule: $\displaystyle \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}$ and $\displaystyle \frac{dX}{dx}= 1$ because X= x- 1.

    Dividing both numerator and denominator of that fraction by X gives $\displaystyle \frac{dY}{dX}= \frac{4-\frac{Y}{X}}{2+ \frac{Y}{X}}$. Let V= Y/X. Then Y= XV and $\displaystyle \frac{dY}{dX}= V+ X\frac{dV}{dX}= \frac{4- V}{2+ V}$. Then $\displaystyle \frac{2+ V}{4- V}\frac{dV}{dX}= dX$.

    First do the "division" indicated by the fraction. 4- V divides into 2+ V negative one times with remainder -2. That is, $\displaystyle \frac{2+ V}{4- V}= -1- \frac{2}{4- V}$. Integrate that with respect to V.
    Last edited by HallsofIvy; May 19th 2018 at 02:34 PM.
    Thanks from s_ingram
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    243
    Thanks
    3

    Re: First order ODE getting to homogenous with x and y substitutions

    Excellent! Many thanks, HallsofIvy, once again! I made a typo, you are right the problem said x = X - 1.
    Last edited by s_ingram; May 20th 2018 at 03:19 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2009
    Posts
    243
    Thanks
    3

    Re: First order ODE getting to homogenous with x and y substitutions

    I think you made a mistake at the end.

    $ X \dfrac{dV}{dX} = \dfrac{4 - V}{2 + V} - V$ which gives $X \dfrac{dV}{dX} = \dfrac{(4 + V)(1 - V)}{(2 + V)}$ but this is just mechanics. Many thanks for setting me on the right path.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,044
    Thanks
    3173

    Re: First order ODE getting to homogenous with x and y substitutions

    Quote Originally Posted by HallsofIvy View Post
    Your original equation is $\displaystyle \frac{dy}{dx}= \frac{4x- y+ 7}{2x+ y- 1}$. You make the substitution x= X- 1 (NOT "x= X+ 1"), y= Y+ 3. $\displaystyle \frac{dy}{dx}= \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}= \frac{dY}{dX}$. 4x-y+ 7= 4(X- 1)- Y- 3+ 7= 4X- Y and 2x+ y- 1= 2(X- 1)+ Y+ 3- 1= 2X+ Y. Yes, that makes the equation $\displaystyle \frac{dY}{dX}= \frac{4X- Y}{2X+ Y}$.

    As to how I went from $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dY}{dx}$, first I replaced y by Y+ 3: $\displaystyle \frac{dY}{dx}= \frac{Y+ 3}{dx}= \frac{dY}{dx}+ 0= \frac{dY}{dx}$. Now to go from $\displaystyle \frac{dY}{dx}$ to $\displaystyle \frac{dY}{dX}$, use the chain rule: $\displaystyle \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}$ and $\displaystyle \frac{dX}{dx}= 1$ because X= x- 1.

    Dividing both numerator and denominator of that fraction by X gives $\displaystyle \frac{dY}{dX}= \frac{4-\frac{Y}{X}}{2+ \frac{Y}{X}}$. Let V= Y/X. Then Y= XV and $\displaystyle \frac{dY}{dX}= V+ X\frac{dV}{dX}= \frac{4- V}{2+ V}$. Then $\displaystyle \frac{2+ V}{4- V}\frac{dV}{dX}= dX$.
    Oh, bother! I did, in fact, forget the "V" term on the left: $\displaystyle X\frac{dV}{dX}= \frac{4- V}{2+ V}- V= \frac{4- V}{2+ V}- \frac{2V+ V^2}{2+ V}= \frac{4- 3V- V^2}{2+ V}= \frac{(4- V)(1+ V)}{2+ V}$.

    $\displaystyle \frac{2+ V}{(4- V)(1+ V)}dV= \frac{1}{X}dX$.

    First do the "division" indicated by the fraction. 4- V divides into 2+ V negative one times with remainder -2. That is, $\displaystyle \frac{2+ V}{4- V}= -1- \frac{2}{4- V}$. Integrate that with respect to V.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    243
    Thanks
    3

    Re: First order ODE getting to homogenous with x and y substitutions

    I get

    $ \dfrac{4 - 3V - V^2}{2 + V} = \dfrac{(4 + V)(1 - V)}{2 + V}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. First order, homogenous DE
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Feb 27th 2013, 05:00 AM
  2. Second Order DE (non homogenous)
    Posted in the Differential Equations Forum
    Replies: 8
    Last Post: Aug 13th 2010, 02:12 PM
  3. second order non-homogenous IVP
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Jul 23rd 2009, 06:44 AM
  4. Third Order Non-Homogenous ODE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jun 25th 2009, 05:08 AM
  5. Replies: 1
    Last Post: May 11th 2007, 04:01 AM

/mathhelpforum @mathhelpforum