Making $\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$ stationary using Euler's equation

Question

9.2.6 - Mary L Boas:

This problem asks us to make the following integral stationary using Euler's equation.

$$\int_{x_1}^{x_2}(y'^2\,+\,\sqrt{y})\,dx$$

If we have $$F=y'^2\,+\,\sqrt{y}$$ then $$\frac{d}{dx}\,\frac {\partial F}{\partial y'}-\frac {\partial F}{\partial y}=0\, \implies \frac{d}{dx}2y'\,-\,\frac{1}{2}y^{-1/2}=0 \, \implies 2y''\,-\,\frac{1}{2}y^{-1/2}=0$$also multiplying by $y'$ and integrating each term we get $$\require{cancel}y'^2\,-\sqrt{y}\,=\,k\implies \int \frac{dy}{(k+\sqrt {y})^\cancelto{1/2}{2}}=x+\mathrm{constant} $$ Here i stumble, because my answer comes nowhere close to the form of given solution: $x+a=\frac{4}{3}(y^{1/2}-2b)(b+y^{1/2})^{1/2}$. Please help.

Edited according to user10354138, now it's ok.

Answer 1

We have $$y'^{2}=k+\sqrt{y}\implies\int\frac{1}{({k+\sqrt{y}})^{\frac{1}{2}}}dy=x+\text{constant}$$

Then substitute $u={k+\sqrt{y}}$ on the LHS so $du=\frac{1}{2\sqrt{y}}dy=\frac{1}{2(u-k)}dy$ and we have

$$x+\text{constant}=\int \frac{2(u-k)}{u^{\frac{1}{2}}}du=2\int u^{\frac{1}{2}}du-2k\int u^{-\frac{1}{2}}du$$ $$=\frac{4}{3}u^{\frac{3}{2}}-4ku^{\frac{1}{2}}=\frac{4}{3}u^{\frac{1}{2}}(u-3k)=\frac{4}{3}(k+y^{\frac{1}{2}})^{\frac{1}{2}}(y^{\frac{1}{2}}-2k)$$

Or $$x+a=\frac{4}{3}(y^{\frac{1}{2}}-2b)(b+y^{\frac{1}{2}})^{\frac{1}{2}}$$

since $a$ and $b$ are constants.

Answer 2

Set $y=u^2$ so that $\sqrt{y}=u$, then $$ 4u^2u'^2-u=k,\\~\\ x+c=\int\frac{2udu}{\sqrt{u+k}} =2\int\sqrt{u+k}\,du-2k\int\frac{du}{\sqrt{k+u}} \\ =\frac43(u+k)^{\frac32}-4k\sqrt{u+k} $$