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Answer to Question #2813 in Differential Equations for Ahmad

Question #2813
Solve y''=(y')2 +y2
Expert's answer
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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