\[ y'(x)-\frac {y(x)-x^2 \sqrt {x^2-y(x)^2}}{x y(x) \sqrt {x^2-y(x)^2}+x}=0 \] ✓ Mathematica : cpu = 2.56867 (sec), leaf count = 44
\[\text {Solve}\left [-\tan ^{-1}\left (\frac {\sqrt {x^2-y(x)^2}}{y(x)}\right )+\frac {x^2}{2}+\frac {y(x)^2}{2}=c_1,y(x)\right ]\] ✓ Maple : cpu = 0.384 (sec), leaf count = 34
\[\frac {y \relax (x )^{2}}{2}+\arctan \left (\frac {y \relax (x )}{\sqrt {x^{2}-y \relax (x )^{2}}}\right )+\frac {x^{2}}{2}-c_{1} = 0\]
Hand solution
\begin {equation} y^{\prime }=\frac {y-x^{2}\sqrt {x^{2}-y^{2}}}{xy\sqrt {x^{2}-y^{2}}+x}\tag {1} \end {equation} Let \(y=ux\) then \(y^{\prime }=u+xu^{\prime }\) therefore\begin {align*} u+xu^{\prime } & =\frac {y-x^{2}\sqrt {x^{2}-y^{2}}}{xy\sqrt {x^{2}-y^{2}}+x}\\ & =\frac {ux-x^{2}\sqrt {x^{2}-\left (ux\right ) ^{2}}}{x\left (ux\right ) \sqrt {x^{2}-\left (ux\right ) ^{2}}+x}\\ & =\frac {ux-x^{3}\sqrt {1-u^{2}}}{x^{3}u\sqrt {1-u^{2}}+x}\\ & =\frac {u-x^{2}\sqrt {1-u^{2}}}{x^{2}u\sqrt {1-u^{2}}+1} \end {align*}
Hence\begin {align*} u\left (x^{2}u\sqrt {1-u^{2}}+1\right ) +xu^{\prime }\left (x^{2}u\sqrt {1-u^{2}}+1\right ) & =u-x^{2}\sqrt {1-u^{2}}\\ x^{2}u^{2}\sqrt {1-u^{2}}+u+u^{\prime }\left (x^{3}u\sqrt {1-u^{2}}+x\right ) & =u-x^{2}\sqrt {1-u^{2}}\\ x^{2}u^{2}\sqrt {1-u^{2}}+u^{\prime }\left (x^{3}u\sqrt {1-u^{2}}+x\right ) & =-x^{2}\sqrt {1-u^{2}}\\ xu^{2}\sqrt {1-u^{2}}+u^{\prime }\left (x^{2}u\sqrt {1-u^{2}}+1\right ) & =-x\sqrt {1-u^{2}}\\ xu^{2}+u^{\prime }\left (x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\right ) & =-x\\ x\left (1+u^{2}\right ) +u^{\prime }\left (x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\right ) & =0 \end {align*}
Hence\begin {equation} x\left (1+u^{2}\right ) dx+\left (x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\right ) du=0\tag {2} \end {equation} Let \(M=x\left (1+u^{2}\right ) ,N=\left (x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\right ) \). \begin {align*} \frac {\partial M}{\partial u} & =2xu\\ \frac {\partial N}{\partial x} & =2xu \end {align*}
Therefore (2) is exact. Let \[ x\left (1+u^{2}\right ) dx+\left (x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\right ) du=dU \] Since \(dU=\frac {\partial U}{\partial x}dx+\frac {\partial U}{\partial u}du\). Comparing with the above, we see that \begin {align} \frac {\partial U}{\partial x} & =x\left (1+u^{2}\right ) \tag {3}\\ \frac {\partial U}{\partial u} & =x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\tag {4} \end {align}
From (3)\begin {align} U & =\int x\left (1+u^{2}\right ) dx\nonumber \\ & =\frac {x^{2}}{2}\left (1+u^{2}\right ) +f\relax (u) \tag {5} \end {align}
From (4)\begin {align*} \frac {d}{\partial u}\left (\frac {x^{2}}{2}\left (1+u^{2}\right ) +f\left ( u\right ) \right ) & =x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\\ x^{2}u+f^{\prime }\relax (u) & =x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\\ f^{\prime }\relax (u) & =\frac {1}{\sqrt {1-u^{2}}} \end {align*}
Therefore \[ f\relax (u) =\arcsin \relax (u) \] From (5) we find\[ U\left (x,u\right ) =\frac {x^{2}}{2}\left (1+u^{2}\right ) +\arcsin \left ( u\right ) \] Since \(dU=0\) then\begin {align*} \frac {x^{2}}{2}\left (1+u^{2}\right ) +\arcsin \relax (u) & =C\\ \frac {x^{2}}{2}\left (1+u^{2}\right ) +\arcsin \relax (u) -C & =0 \end {align*}
Since \(y=ux\) then the above can be written as\begin {align*} \frac {x^{2}}{2}\left (1+\left (\frac {y}{x}\right ) ^{2}\right ) +\arcsin \left (\frac {y}{x}\right ) -C & =0\\ \frac {x^{2}}{2}\left (\frac {x^{2}+y^{2}}{x^{2}}\right ) +\arcsin \left ( \frac {y}{x}\right ) -C & =0\\ \frac {1}{2}\left (x^{2}+y^{2}\right ) +\arcsin \left (\frac {y}{x}\right ) -C & =0\\ \arcsin \left (\frac {y}{x}\right ) & =C-\frac {1}{2}\left (x^{2}+y^{2}\right ) \end {align*}
Hence\begin {align*} \frac {y}{x} & =\sin \left (C-\frac {1}{2}\left (x^{2}+y^{2}\right ) \right ) \\ y\relax (x) & =x\sin \left (C-\frac {1}{2}\left (x^{2}+y^{2}\right ) \right ) \end {align*}