#### 2.115   ODE No. 115

$-x (y(x)-x) \sqrt {x^2+y(x)^2}+x y'(x)-y(x)=0$ Mathematica : cpu = 0.175946 (sec), leaf count = 221

$\left \{\left \{y(x)\to \frac {x-2 \sqrt {x^2 \tanh ^2\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )-x^2 \tanh ^4\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )}}{2 \tanh ^2\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )-1}\right \},\left \{y(x)\to \frac {2 \sqrt {x^2 \tanh ^2\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )-x^2 \tanh ^4\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )}+x}{2 \tanh ^2\left (\frac {1}{2} \left (-2 \sqrt {2} c_1-\sqrt {2} x^2\right )\right )-1}\right \}\right \}$ Maple : cpu = 0.171 (sec), leaf count = 49

$\left \{ \ln \left ( 2\,{\frac {x \left ( \sqrt {2\, \left ( y \left ( x \right ) \right ) ^{2}+2\,{x}^{2}}+y \left ( x \right ) +x \right ) }{y \left ( x \right ) -x}} \right ) +{\frac {\sqrt {2}{x}^{2}}{2}}-\ln \left ( x \right ) -{\it \_C1}=0 \right \}$

Hand solution

$xy^{\prime }=x\left ( y-x\right ) \sqrt {y^{2}-x^{2}}+y$

Let $$y=xu$$, then $$y^{\prime }=u+xu^{\prime }$$ and the above becomes

\begin {align*} x\left ( u+xu^{\prime }\right ) & =x\left ( xu-x\right ) \sqrt {\left ( xu\right ) ^{2}-x^{2}}+xu\\ \left ( u+xu^{\prime }\right ) & =\left ( xu-x\right ) \sqrt {\left ( xu\right ) ^{2}-x^{2}}+u\\ xu^{\prime } & =\left ( xu-x\right ) x\sqrt {u^{2}-1}\\ u^{\prime } & =x\left ( u-1\right ) \sqrt {u^{2}-1} \end {align*}

Separable.

\begin {align*} \frac {du}{\left ( u-1\right ) \sqrt {u^{2}-1}} & =xdx\\ \frac {-u-1}{\sqrt {u^{2}-1}} & =\frac {x^{2}}{2}+C \end {align*}

But $$y=xu$$, hence

$\frac {-\frac {y}{x}-1}{\sqrt {\left ( \frac {y}{x}\right ) ^{2}-1}}=\frac {x^{2}}{2}+C$

Let $$\frac {y}{x}=z$$

\begin {align*} \frac {-z-1}{\sqrt {z^{2}-1}} & =\frac {x^{2}}{2}+C\\ -z-1 & =\sqrt {z^{2}-1}\left ( \frac {x^{2}}{2}+C\right ) \\ \left ( -z-1\right ) ^{2} & =\left ( z^{2}-1\right ) \left ( \frac {x^{2}}{2}+C\right ) ^{2}\\ z^{2}+1+2z & =z^{2}\left ( \frac {x^{2}}{2}+C\right ) ^{2}-\left ( \frac {x^{2}}{2}+C\right ) ^{2}\\ z^{2}\left ( 1-\left ( \frac {x^{2}}{2}+C\right ) ^{2}\right ) +2z+1+\left ( \frac {x^{2}}{2}+C\right ) ^{2} & =0 \end {align*}

$$\allowbreak$$

Solving for $$z$$ (quadratic formula, some conditions apply), one of the solutions is

$z=\frac {4Cx^{2}+4C^{2}+x^{4}+4}{4Cx^{2}+4C^{2}+x^{4}-4}$

Hence

$y=x\frac {4Cx^{2}+4C^{2}+x^{4}+4}{4Cx^{2}+4C^{2}+x^{4}-4}$

Need to work on veriﬁcation. Kamke gives the ﬁnal solution as

$y=x\frac {-2Cx^{2}+C^{2}+x^{4}+4}{-2Cx^{2}+C^{2}+x^{4}-4}$

I am not sure where my error now is. Need to look at this again.