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Question #2813
Solve y''=(y')[sup]2[/sup] +y[sup]2[/sup]
Assume y&#039;=p, then y&#039;&#039;=pp&#039;. Let&rsquo;s substitute these variables into equation:
pp&#039;=p2+y2
1/2 (p2 )&#039;-p2=y2

Let&rsquo;s solve the last equation with respect to p2=f(y) as a function of y:
f&#039;-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&#039;=2ay+b
f&#039;-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:
The last equation hasn&rsquo;t analytical solution.

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