#### 2.112   ODE No. 112

$-\sqrt {x^2+y(x)^2}+x y'(x)-y(x)=0$ Mathematica : cpu = 0.0981619 (sec), leaf count = 13

$\{\{y(x)\to x \sinh (c_1+\log (x))\}\}$ Maple : cpu = 0.03 (sec), leaf count = 27

$\left \{ {\frac {1}{{x}^{2}}\sqrt { \left ( y \left ( x \right ) \right ) ^{2}+{x}^{2}}}+{\frac {y \left ( x \right ) }{{x}^{2}}}-{\it \_C1}=0 \right \}$

Hand solution

$xy^{\prime }=\sqrt {x^{2}+y^{2}}+y$

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

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

Separable.

$\frac {dv}{\sqrt {1+v^{2}}}=\frac {1}{x}dx$

Integrating

\begin {align*} \operatorname {arcsinh}\left ( v\right ) & =\ln x+C\\ v & =\sinh \left ( \ln x+C\right ) \end {align*}

Since $$y=xv$$ then

$y=x\sinh \left ( \ln x+C\right )$

Veriﬁcation

ode:=x*diff(y(x),x)=sqrt(x^2+y(x)^2)+y(x);
y0:=x*sinh(ln(x)+_C1);
odetest(y(x)=y0,ode) assuming x>= 0;
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