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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 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: 2,256 Thanks: 926 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: 7,674 Thanks: 2654 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: 7,674 Thanks: 2654 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: 7,674 Thanks: 2654 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: 7,674 Thanks: 2654 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: 7,674 Thanks: 2654 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 Tags challenge, ode Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Denis New Users 9 August 1st, 2013 01:45 PM mathmaniac New Users 5 March 12th, 2013 03:16 PM mattpi Number Theory 2 September 30th, 2009 02:39 PM Challenger Advanced Statistics 1 January 27th, 2009 03:34 PM mattpi New Users 0 December 31st, 1969 04:00 PM

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