May 13th, 2015, 05:28 AM  #1 
Newbie Joined: May 2015 From: Israel Posts: 6 Thanks: 0  ODE  third challenge
So i'm stucked with this equation, and need help.. i've started solve it but than it arrived to an integral of 1/lnc1y which i don't know how to solve  the equation is: 1/y(y'^2)y''+2y''^2+y'y'''=0 thanks a lot 
May 13th, 2015, 05:40 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,041 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff. 
Where did you get this monstrosity from? Context might be helpful. Dan 
May 13th, 2015, 06:08 AM  #3 
Newbie Joined: May 2015 From: Israel Posts: 6 Thanks: 0 
it's from our work we have to give till tomorrow, it means i need to decrease the equation from 3 order till 1 order.. than i used: y=p' y''=p'p y'''=(p''p^2+p'^2p) than i used the fact that: p'/p = (lnp)'= z and than p=e^integral of zdy p'=ze^integral of zdy p''=(z'e^integralofzdy+z^2e^integralofzdy) 
May 13th, 2015, 08:11 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
Is that $${1 \over y}(y'^2)y''+2y''^2+y'y'''=0$$ or $${1 \over y(y'^2)y''}+2y''^2+y'y'''=0$$ I assume the first form. 
May 13th, 2015, 08:32 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
I don't agree with your derivatives of $y=p'$. I notice that: $$\newcommand{\ddx}{{\mathrm d \over \mathrm d x}} \ddx yy'y'' = y'^2y'' + y\ddx y'y''= y'^2y'' + y(y''^2 + y'y''')$$ So we have $${1 \over y}\ddx yy'y'' + y''^2 = 0$$ I'm not sure that is helps though other than that $y = c \ne 0$, a constant is clearly a solution. 
May 13th, 2015, 08:33 AM  #6  
Newbie Joined: May 2015 From: Israel Posts: 6 Thanks: 0  Quote:
thanks a lot! if you would like to help me with another equation  this is a cauchy problem: y'y'''3(y'')^2((2y')^5)/y^3=0 y(0)=y'(0)=1 y''(0)=1 thanks a lot! you're awesome!  
May 13th, 2015, 08:48 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
Is that really $(2y')^5 = 32y'^5$?

May 13th, 2015, 09:06 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
$$\newcommand{\ddx}{{\mathrm d \over \mathrm d x}} y'y'''3y''^2 {2y'^5 \over y^3} = y'^2 \left( {y'y'''(y'')^2 \over y'^2} 2\left({y''^2 \over y'^2} + {y'^3 \over y^3}\right) \right) = y'^2 \left(\ddx {y'' \over y'} 2\left({y''^2 \over y'^2} + {y'^3 \over y^3}\right) \right)=0$$ Again, I'm not sure how useful that is. 
May 13th, 2015, 09:47 AM  #9 
Newbie Joined: May 2015 From: Israel Posts: 6 Thanks: 0 
i dont know either.. but thank you for being helpful

May 13th, 2015, 10:08 AM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra 
Since $y'(0) = 1 \ne 0$ it is not identically zero, so we can say that the expression in brackets is identically zero. Integrating then gives $${y'' \over y'}  2\log {y''^2y' \over y^3} = c$$ 

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