Assume y'=p, then y''=pp'. Let’s substitute these variables into equation: pp'=p2+y2 1/2 (p2 )'-p2=y2 Let’s solve the last equation with respect to p2=f(y) as a function of y: f'-2f=2y2 Firstly find the general solution of the homogeneous equation: t - 2=0 t=2 f=C1 e2y Then find the particular solution of the non-homogeneous equation: f=ay2+by+c f'=2ay+b f'-2f=2ay+b - 2(ay2+by+c)=-2ay2+(2a-2b)y+(b-2c) = 2y2
The following system of equations -2a=2 2a-2b=0 b-2c=0 has roots: (a=-1, b=-1, c=-1/2)
Thus, the general solution of the non-homogeneous equation is: f=C1 e2y - y2- y - 1/2 Returning to the substitution: y' = √(p2 ) = √f = √(C1e2y - y2- y - 1/2) The last equation hasn’t analytical solution.
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