# Answer to Question #2783 in Differential Equations for Ahmad

Question #2783

solve my differential equation

y'' =(y')

y'' =(y')

^{3}+ x^{2}Expert's answer

Let’s solve the homogeneous equation:

Denote y' as p:

Let’s solve nonhomogeneous equitation:

The last equitation hasn’t analytical solution.

**y''- y'**^{3}= 0Denote y' as p:

**p'p - p**

dp/dy = p

p = -1/x + C^{3}= 0dp/dy = p

^{2}p = -1/x + C

Let’s solve nonhomogeneous equitation:

**p'p - p**

p'p

dp/dx = 1/(p (C - p)

dx = (p(C - p)^{3}= 1/p^{2}p'p

^{3}- p^{5}= 1dp/dx = 1/(p (C - p)

^{2}) + p^{2}dx = (p(C - p)

^{2})/(1 + p^{3}(C - p)^{2}) dpThe last equitation hasn’t analytical solution.

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