# Easy way to solve this non-linear second degree DY $\sqrt{y}\;y''=1$?

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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Clever idea +1! – hhh Feb 17 '12 at 14:08
@hhh,You can use this method whenever $x$ doesn't appear explicitly in the equation . – pedja Feb 17 '12 at 14:18

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

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

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

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

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

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

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

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

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