# Answer to Question #2813 in Differential Equations for Ahmad

Question #2813

Solve y''=(y')

^{2}+y^{2}Expert's answer

Assume y'=p, then y''=pp'. Let’s substitute these variables into equation:

Let’s solve the last equation with respect to p

Firstly find the general solution of the homogeneous equation:

Then find the particular solution of the non-homogeneous equation:

f

The following system of equations

has roots: (a=-1, b=-1, c=-1/2)

Thus, the general solution of the non-homogeneous equation is:

f

Returning to the substitution:

The last equation hasn’t analytical solution.

**pp'=p**

1/2 (p^{2}+y^{2}1/2 (p

^{2})'-p^{2}=y^{2}Let’s solve the last equation with respect to p

^{2}=f(y) as a function of y:**f'-2f=2y**^{2}Firstly find the general solution of the homogeneous equation:

**t - 2=0 t=2**

f=Cf=C

_{1}e^{2y}Then find the particular solution of the non-homogeneous equation:

f

**=ay**

f'=2ay+b

f'-2f=2ay+b - 2(ay^{2}+by+cf'=2ay+b

f'-2f=2ay+b - 2(ay

^{2}+by+c)=-2ay^{2}+(2a-2b)y+(b-2c) = 2y^{2}The following system of equations

**-2a=2**

2a-2b=0

b-2c=02a-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

**=C**_{1}e^{2y}- y^{2}- y - 1/2Returning to the substitution:

**y' = √(p**^{2}) = √f = √(C_{1}e^{2y}- y^{2}- y - 1/2)The last equation hasn’t analytical solution.

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