Differential calculus, solve for y: 4(y''y'')+(y'y')-1=0

[endykami]

Homework Statement

quadratic differential calculus

Relevant Equations

4(y''y'')+(y'y')-1=0
solve for y

suppose y''=r^2=s
y'=r
4(y''y'')-(y'y')-1=0=4(r^2)^2-(r^2)-1=4(s^2)-s-1
s=(-b±√(b^2-4ac))/2a
s=(1±√17)/8
y=∫∫sdx=∫∫((1±√17)/8) dx=(1±√17)/8)(1/2)x^2+c1x+c2

 

[BvU]

Hello endykami $\quad$ $\quad$ !

suppose y''=r^2=s
y'=r

If y'=r then y'' = r', not r²

 

[epenguin]

You might as well define a new variable e.g. r, let y' = r to recognise you really ha e a first order d.e. To solve in the first place.
Are you sure you transcribed the problem right? - it is a bit unusual to write all squares as (x.x)

 

[benorin]

Here, I’ll get you started: let y’=r => y’’=r’ and the equation becomes:
4(r’)^2-r^2+1=0