Solution of the differential equation $\frac{dy}{dx}=\sqrt{\frac{x}{1-x}}$ is a family of

Question

Solution of the differential equation $\frac{dy}{dx}=\sqrt{\frac{x}{1-x}}$ is a family of

(a) hyperbolas (b) ellipses (c) parabolas (d) concentric circles

After solving the integration, i obtained $y= \sin^{-1}(\sqrt{x})-\sqrt{x}\sqrt{1-x}+c$. Which option should be correct?

Answer 1

I think that your differential equation is not correct. $$\dfrac {d}{dx}\dfrac {\partial F}{ \partial y'}-\dfrac {\partial F}{ \partial y}=0$$ Since $F=F(x,y')$ we have: $$\dfrac {\partial F}{ \partial y'}=C$$ $$\dfrac {y'\sqrt x }{ \sqrt {1+y'^2}}=C$$ $$y'^2(x-C^2)=C^2$$ $$y'=\pm \dfrac {C}{\sqrt {x-C^2}}$$ $$y+A=\pm \int \dfrac {Cdx}{\sqrt {x-C^2}}$$ $$(y+A)^2=4C^2( {x-C^2})$$ I think you should end with a family of parabolas. $$Y^2=2pX$$

Answer 2

The requirement for the equation to even make sense conceptually is that $$\frac{x}{1-x}\geq0$$ and $x\neq1.$ The former is equivalent to $$\frac{-x}{1-x}=\frac{1-x-1}{1-x}=1-\frac1{1-x}\leq0,$$ which is equivalent to $$\frac1{1-x}\geq1,$$ which is equivalent to $0\lt1-x\leq1,$ which is equivalent to $-1\leq{x-1}\lt0,$ equivalent to $0\leq{x}\lt1.$ With this in mind, the equation proposed is $$y'(x)=\sqrt{\frac{x}{1-x}}.$$ By the fundamental theorem of calculus, $$\int_0^ty'(x)\,\mathrm{d}x=y(t)-y(0),$$ hence $$y(t)=y(0)+\int_0^t\sqrt{\frac{x}{1-x}}\,\mathrm{d}x.$$ Notice that $$\int_0^t\sqrt{\frac{x}{1-x}}\,\mathrm{d}x=\int_0^tx^{\frac12}(1-x)^{-\frac12}\,\mathrm{d}x=\int_0^tx^{\frac32-1}(1-x)^{\frac12-1}\,\mathrm{d}x=\mathrm{B}\left(t;\frac32,\frac12\right),$$ where $$\mathrm{B}(t;a,b)=\int_0^tx^{a-1}(1-x)^{b-1}\,\mathrm{d}x$$ is the incomplete Beta function.

Also, notice that $$\int_0^t\sqrt{\frac{x}{1-x}}\,\mathrm{d}x=\int_0^t\frac{\sqrt{x}}{\sqrt{1-\sqrt{x}^2}}\,\mathrm{d}x=\int_0^t\frac{2\sqrt{x}^2}{\sqrt{1-\sqrt{x}^2}}\frac1{2\sqrt{x}}\,\mathrm{d}x,$$ so with $y=\sqrt{x},$ one has that $$\int_0^t\sqrt{\frac{x}{1-x}}\,\mathrm{d}x=\int_0^{\sqrt{t}}\frac{2y^2}{\sqrt{1-y^2}}\,\mathrm{d}y=\int_0^{\sqrt{t}}\frac{2y^2-2+2}{\sqrt{1-y^2}}\,\mathrm{d}y=\int_0^{\sqrt{t}}-2\sqrt{1-y^2}+\frac2{\sqrt{1-y^2}}\,\mathrm{d}y=2\arcsin(\sqrt{t})-2\int_0^{\sqrt{t}}\sqrt{1-y^2}\,\mathrm{d}y.$$ Remember that $$\int_0^{\sqrt{t}}\sqrt{1-y^2}\,\mathrm{d}y=\sqrt{t}\sqrt{1-t}-\int_0^{\sqrt{t}}-\frac{y^2}{\sqrt{1-y^2}}\,\mathrm{d}y=\sqrt{t}\sqrt{1-t}+\int_0^{\sqrt{t}}\frac{y^2}{\sqrt{1-y^2}}\,\mathrm{d}y.$$ Thus $$2\int_0^{\sqrt{t}}\frac{y^2}{\sqrt{1-y^2}}\,\mathrm{d}y=2\arcsin(\sqrt{t})-2\sqrt{t}\sqrt{1-t}-2\int_0^{\sqrt{t}}\frac{y^2}{\sqrt{1-y^2}}\,\mathrm{d}y,$$ which is equivalent to $$\int_0^{\sqrt{t}}\frac{2y^2}{\sqrt{1-y^2}}\,\mathrm{d}y=\arcsin(\sqrt{t})-\sqrt{t}\sqrt{1-t}.$$ Therefore, $$y(t)=y(0)+\arcsin(\sqrt{t})-\sqrt{t}\sqrt{1-t},$$ which indeed, verifies your solution of the equation is correct. However, this is not a conic section. So your differential equation must actually be incorrect, if the solution is supposed to be a conic section.