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I prooceeded by integrating both sides y'=\int y^{-\frac{1}{2}} dx=\cdots

so I got (y')^{2}+C y' - \frac{1}{2} \sqrt{y} = 0 but I am thinking that I am proceeding the wrong or the hard way. Some easy to solve this kind of 2nd-degree DYs?

Page 633 on the book I have been reading earlier.

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Substitute : y_x' = v \Rightarrow y_x''=v_y' \cdot v~ , hence :

\sqrt {y} \cdot v_y' \cdot v=1 \Rightarrow \int v \,dv = \int \frac{dy}{\sqrt{y}}

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    @hhh,You can use this method whenever $x$ doesn't appear explicitly in the equation .2012-02-17
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y'=\int y^{-\frac{1}{2}} dx=...

y''= y^{-\frac{1}{2}} y'y''= y'y^{-\frac{1}{2}}

\int y'y'' dx=\int y^{-\frac{1}{2}} y'dx

\frac {y'^{2}}{2} = 2y^{\frac{1}{2}} +k

y'^{2} = 4y^{\frac{1}{2}} +2k=4y^{\frac{1}{2}} +c

y' = \sqrt{4y^{\frac{1}{2}} +c}

\frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} = 1 \int \frac{y'}{\sqrt{4y^{\frac{1}{2}} +c}} dx=\int dx

If you select $ u^{2}= 4y^{\frac{1}{2}} +c $

$ y= \frac{(u^{2}-c)^{2}}{16} $

y'= uu'\frac{(u^{2}-c)}{4}

\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=\int dx

\int \frac{u'u\frac{(u^{2}-c)}{4}}{u} dx=x+c_1

\int \frac{u'u^{2}}{4} dx -\int \frac{u'c}{4} dx=x+c_1

$\frac{u^{3}}{12} -\frac{cu}{4} =x+c_1$

After solving cubic equation you must put $ u= \sqrt{4y^{\frac{1}{2}} +c} $

then you must find y depend on X