Answer to Question #2813 in Differential Equations for Ahmad

Question

Solve y''=(y')[sup]2[/sup] +y[sup]2[/sup]

Answer 1

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 e^2y
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 e^2y - y²- y - 1/2
Returning to the substitution:
y' = √(p² ) = √f = √(C₁e^2y - y²- y - 1/2)
The last equation hasn’t analytical solution.