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 May 13th, 2015, 04:28 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, 04:40 AM #2 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. Where did you get this monstrosity from? Context might be helpful. -Dan
 May 13th, 2015, 05:08 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, 07:11 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 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, 07:32 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 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, 07:33 AM   #6
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Joined: May 2015
From: Israel

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Quote:
 Originally Posted by v8archie 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.
indeed it's the first one, but it's irrelevant because my lecturer just told us there is a mistake with this exercise.
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, 07:48 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra Is that really $(2y')^5 = 32y'^5$?
 May 13th, 2015, 08:06 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 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. Thanks from lola94
 May 13th, 2015, 08:47 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, 09:08 AM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 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 $${y'' \over y'} - 2\log {y''^2y' \over y^3} = c$$ Thanks from lola94

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