2.62 ODE No. 62
\[ y'(x)-\frac {y(x)-x^2 \sqrt {x^2-y(x)^2}}{x y(x) \sqrt {x^2-y(x)^2}+x}=0 \]
✓ Mathematica : cpu = 1.72298 (sec), leaf count = 44
DSolve[-((y[x] - x^2*Sqrt[x^2 - y[x]^2])/(x + x*y[x]*Sqrt[x^2 - y[x]^2])) + Derivative[1][y][x] == 0,y[x],x]
\[\text {Solve}\left [-\tan ^{-1}\left (\frac {\sqrt {x^2-y(x)^2}}{y(x)}\right )+\frac {x^2}{2}+\frac {y(x)^2}{2}=c_1,y(x)\right ]\]
✓ Maple : cpu = 0.503 (sec), leaf count = 34
dsolve(diff(y(x),x)-(y(x)-x^2*(x^2-y(x)^2)^(1/2))/(x*y(x)*(x^2-y(x)^2)^(1/2)+x) = 0,y(x))
\[\frac {y \left (x \right )^{2}}{2}+\arctan \left (\frac {y \left (x \right )}{\sqrt {x^{2}-y \left (x \right )^{2}}}\right )+\frac {x^{2}}{2}-c_{1} = 0\]
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\left ( u\right ) \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 }\left ( u\right ) & =x^{2}u+\frac {1}{\sqrt {1-u^{2}}}\\ f^{\prime }\left ( u\right ) & =\frac {1}{\sqrt {1-u^{2}}}\end{align*}
Therefore
\[ f\left ( u\right ) =\arcsin \left ( u\right ) \]
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 \left ( u\right ) & =C\\ \frac {x^{2}}{2}\left ( 1+u^{2}\right ) +\arcsin \left ( u\right ) -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\left ( x\right ) & =x\sin \left ( C-\frac {1}{2}\left ( x^{2}+y^{2}\right ) \right ) \end{align*}