Question #2813

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

Expert's answer

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

**pp'=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=C_{1} e^{2y}

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

f**=ay**^{2}+by+c

f'=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=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 - 1/2

Returning to the substitution:

**y' = √(p**^{2} ) = √f = √(C_{1}e^{2y} - y^{2}- y - 1/2)

The last equation hasn’t analytical solution.

1/2 (p

Let’s solve the last equation with respect to p

Firstly find the general solution of the homogeneous equation:

f=C

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

f

f'=2ay+b

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

The following system of equations

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

Returning to the substitution:

The last equation hasn’t analytical solution.

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