Reduction of order (II)

Problem:

$ y'' - y'y = 0 $

Solution:

This time we see $ x $ does not appear in the equation, we can let $ z = y' $ and $ y'' = \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} = z'z $, note that $ z' $ denote $ \frac{dz}{dy} $.

The equation becomes

$ \begin{eqnarray*} y'' - y'y &=& 0 \\ z'z - z y &=& 0 \\ z'z &=& z y \\ z' &=& y \\ z &=& \frac{y^2}{2} + c_1 \\ y' &=& \frac{y^2}{2} + c_1 \\ \frac{2dy}{y^2 + c_2^2} &=& dx \\ 2 \arctan(y/c_2)/c_2 &=& x + c_3 \\ \arctan(y/c_2) &=& \frac{c_2x + c_4}{2} \\ y/c_2 &=& \tan\frac{c_2x + c_4}{2} \\ y &=& c_2 \tan\frac{c_2x + c_4}{2} \\ y &=& 2 c_5 \tan(c_5x + c_6) \\ \end{eqnarray*} $

Check:

$ \begin{eqnarray*} y &=& 2 c_5 \tan(c_5x + c_6) \\ y' &=& 2 c_5 \sec^2(c_5x + c_6) (c_5) \\ &=& 2 c_5^2 \sec^2(c_5x + c_6) \\ y'' &=& 2 c_5^2 (2 \sec(c_5x + c_6))(\sec(c_5x + c_6) \tan(c_5x + c_6)) (c_5) \\ &=& 4 c_5^3 \sec^2(c_5x + c_6) \tan(c_5x + c_6) \end{eqnarray*} $