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: Seconed order non linear ODE

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex] yy''=y'^{2}-y'^{3} [/tex]

    I'm quite sure I got lost somewhere. Can anyone show me where?
    Thanks

    Set

    [tex]z(y)=y' [/tex]

    then

    [tex]\frac{\partial z}{\partial y}=y''\cdot y'=zy'' so y''=\frac{z'}{z} [/tex]

    Plugging this in

    [tex] y\frac{z'}{z}=z^{2}\left(1-z\right) and so \frac{1}{y}\partial y=\frac{\partial z}{z^{3}\left(1-z\right)} [/tex]

    Integrating we have

    [tex] ln\left(y\right)=\int\frac{1}{z^{3}}+\frac{1}{z^{2}}+\frac{1}{z}-\frac{1}{z-1}=-\frac{1}{z^{2}}-\frac{1}{z}+ln\left(\frac{z}{z-1}\right)+c [/tex]

    So

    [tex] y=c_{1}\left(\frac{z}{z-1}\right)e^{-\left(\frac{z+1}{^{z^{2}}}\right)} [/tex] and recalling that z=y' we have

    [tex] y=c_{1}\left(\frac{y'}{y'-1}\right)e^{-\left(\frac{y'+1}{y'^{2}}\right)} [/tex]

    What now?


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2011 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi talolard! :smile:
    no :confused:

    dz/dy = dz/dt dt/dy = y'' / y'​

    (same as a = dv/dt = v dv/dx :wink:)
     
  4. Jun 23, 2011 #3
    I wish I had asked sooner.
    Thanks
    Tal
     
  5. Jun 23, 2011 #4
    [tex]yy''=y'^{2}-y'^{3} [/tex]

    Solution

    Set

    [tex] z(y)=y' [/tex]

    then

    [tex] \frac{\partial z}{\partial y}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial y}=y''\cdot\frac{1}{y'}=y''\frac{1}{z}\rightarrow z\cdot z'=y'' [/tex]

    Plugging this in and assuming z\neq0,1

    [tex] yz\cdot z'=z^{2}\left(1-z\right)\iff yz'=z\left(1-z\right)\iff\frac{\partial z}{z\left(1-z\right)}=\frac{\partial y}{y} [/tex]

    Integrating we have

    [tex] ln\left(y\right)=ln\left(\frac{z}{1-z}\right)+c\iff y=c_{1}\left(\frac{z}{1-z}\right)=c_{1}\left(\frac{y'}{1-y'}\right) [/tex]

    And here it gets tricky because i don't see a nice way to solve this. The best way I found is

    [tex] \left(1-y'\right)=c_{1}\frac{y'}{y}=c_{1}\left(\ln\left(y\right)\right)'\iff c_{1}\left(lny\right)=x-y+c_{2} [/tex]

    Exponentiating we have

    [tex] c_{1}y=c_{2}e^{x}e^{-y}\iff e^{y}y=\frac{c_{1}}{c_{2}}e^{x} [/tex] where ignored the change in the constant.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook