Find this ODE solution $(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$ Find all solutions $y$ to the ODE
$$(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$$
My try:
$$(y-x)\dfrac{d(\sqrt{1+y^2})}{y}=\dfrac{(1+y^2)dx}{\sqrt{x^2+1}}$$
then
$$\dfrac{(y-x)\cdot d(\sqrt{1+y^2})}{y(1+y^2)}=\dfrac{d(\sqrt{1+x^2})}{x}$$
then I can't.Thank you for you help 
 A: Maple says:
$$
\arctan \left( x \right) -\int ^{-\arctan \left( x \right) +\arctan
 \left( y \left( x \right)  \right) }\!{\frac {2-\cos \left( 4\,t+4
 \right) -2\,\cos \left( 2\,t \right) +2\,\cos \left( 2\,t+4 \right) -
\cos \left( 4 \right) +\sqrt {4\,\cos \left( 4\,t+8 \right) -2\,\cos
 \left( 8+2\,t \right) -2\,\cos \left( 6\,t+8 \right) -8\,\cos \left( 
4 \right) +16\,\cos \left( 2\,t+4 \right) -12\,\cos \left( 2\,t
 \right) -8\,\cos \left( 4\,t+4 \right) +12}}{2+\cos \left( 4\,t+4
 \right) +2\,\cos \left( 2\,t \right) +2\,\cos \left( 2\,t+4 \right) +
\cos \left( 4 \right) }}{dt}-C=0
$$
A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
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 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\pars{y - x}\root{x^{2} + 1}\,\totald{y}{x} = \pars{1 + y^{2}}^{3/2}}$

With $\ds{x \equiv \tan\pars{\alpha}}$ and $\ds{y \equiv \tan\pars{\beta}}$ we get:
  $$\!\!\!\!\!
\bracks{\tan\pars{\beta} - \tan\pars{\alpha}}\sec\pars{\alpha}\,
{\sec^{2}\pars{\beta} \over \sec^{2}\pars{\alpha}}\,\totald{\beta}{\alpha}
=\sec^{3}\pars{\beta}\ \imp\
\bracks{\tan\pars{\beta} - \tan\pars{\alpha}}\cos\pars{\alpha}\cos\pars{\beta}\,
\totald{\beta}{\alpha} = 1
$$ 

$$
\bracks{\sin\pars{\beta}\cos\pars{\alpha} - \sin\pars{\alpha}\cos\pars{\beta}}\,
\totald{\beta}{\alpha}= 1\quad\imp\quad
\totald{\beta}{\alpha} = {1 \over \sin\pars{\beta - \alpha}}
$$

With $u = \beta - \alpha$
  $$
\totald{u}{\alpha} + 1 = {1 \over \sin\pars{u}}\quad\imp\quad
\bracks{-1 + {1 \over 1 - \sin\pars{u}}}\,\totald{u}{\alpha} = 1
$$

$$
-u + \int{\dd u \over 1 - \sin\pars{u}} = \alpha + C\,,\qquad C\ \mbox{is a constant}
$$

Also, with $t = \tan\pars{u/2}$:
  \begin{align}
\int{\dd u \over 1 - \sin\pars{u}}&=
\int{1 \over 1 - 2t/\pars{1 + t^{2}}}\,{2\,\dd t \over t^{2} + 1}
=2\int{\dd t \over \pars{t - 1}^{2}} = -\,{2 \over t - 1}
={2 \over 1 - \tan\pars{u/2}}
\\[3mm]&=2\,{\cos\pars{u/2} \over \cos\pars{u/2} - \sin\pars{u/2}}
={2\cos^{2}\pars{u/2} + 2\cos\pars{u/2}\sin\pars{u/2} \over \cos\pars{u}}
\\[3mm]&={1 + \cos\pars{u} + \sin\pars{u} \over \cos\pars{u}}
=\sec\pars{u} + 1 + \tan\pars{u}
\end{align}

Then,
$$
-\beta + \sec\pars{\beta - \alpha} + \tan\pars{\beta - \alpha} = \mbox{constant}
$$
which is an implicit relation between $\alpha = \arctan\pars{x}$ and
$\beta = \arctan\pars{y}$:
$$
-\arctan\pars{y} + {\sqrt{\pars{y - x}^{2} + \pars{1 + yx}^{2}} \over \verts{1 + yx}}
+ {y - x \over 1 + yx} = \mbox{constant}
$$
A: Another CAS gave this "nice" expression 
$$x(y)=\frac{-2 c_1 y+2 c_1 \tan ^{-1}(y)-c_1^2-\tan ^{-1}(y)^2+2 y \tan ^{-1}(y)+1}{c_1^2
   y-2 c_1 y \tan ^{-1}(y)-2 c_1-y+y \tan ^{-1}(y)^2+2 \tan ^{-1}(y)}$$ which simplifies if $y(0)=0$ but does not change to be more pleasant.
