Second order linear: 2 t^2 y" + (y')^3 = 2ty', t > 0

[platinumgold]

\(\displaystyle 31.\, 2t^2 y''\, +\, (y')^3\, =\, 2ty',\, t\, >\, 0\)

\(\displaystyle 31.\, y\, =\, \pm\, \frac{2}{3}\, (t\, -\, 2c_1)\, \sqrt{\strut t\, +\, c_1\,}\, +\, c_2;\, \mbox{ also }\, y\, =\, c\)
. ..\(\displaystyle \mbox{factor.}\)

I tried to solve it. Unfortunately it didn't work. So, please help.

 

[stapel]

\(\displaystyle 31.\, 2t^2 y''\, +\, (y')^3\, =\, 2ty',\, t\, >\, 0\)

\(\displaystyle 31.\, y\, =\, \pm\, \frac{2}{3}\, (t\, -\, 2c_1)\, \sqrt{\strut t\, +\, c_1\,}\, +\, c_2;\, \mbox{ also }\, y\, =\, c\)
. ..\(\displaystyle \mbox{factor.}\)

I tried to solve it. Unfortunately it didn't work.

Please reply showing your efforts, so we can see where things might be going wrong. Thank you!

 

[HallsofIvy]

Perhaps your difficulty lies in your title- you appear to think this is a linear equation. It isn't because of the "(y')^3" term. However, notice that there are y'' and y' terms but no "y". If we let u= y' this becomes the first order equation 2t^2 u'+ u^3= 2tu so that u'= (2tu- u^3)/2t^2 which can be written in "differential form" as 2t^2 du+ (u^3- 2tu)dt= 0. That is not exact but perhaps you can find an integrating factor.