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May 13th, 2015, 04:28 AM   #1
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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
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May 13th, 2015, 04:40 AM   #2
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Where did you get this monstrosity from? Context might be helpful.

-Dan
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May 13th, 2015, 05:08 AM   #3
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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)
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May 13th, 2015, 07:11 AM   #4
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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.
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May 13th, 2015, 07:32 AM   #5
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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.
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May 13th, 2015, 07:33 AM   #6
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Quote:
Originally Posted by v8archie View Post
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!
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May 13th, 2015, 07:48 AM   #7
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Is that really $(2y')^5 = 32y'^5$?
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May 13th, 2015, 08:06 AM   #8
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$$\newcommand{\ddx}{{\mathrm d \over \mathrm d x}}
y'y'''-3y''^2- {2y'^5 \over y^3} = y'^2 \left( {y'y'''-(y'')^2 \over y'^2} -2\left({y''^2 \over y'^2} + {y'^3 \over y^3}\right) \right) = y'^2 \left(\ddx {y'' \over y'} -2\left({y''^2 \over y'^2} + {y'^3 \over y^3}\right) \right)=0$$
Again, I'm not sure how useful that is.
Thanks from lola94
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May 13th, 2015, 08:47 AM   #9
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i dont know either.. but thank you for being helpful
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May 13th, 2015, 09:08 AM   #10
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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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