# Solve the DE $yy^{\prime}+x=\sqrt{x^2+y^2}$

Solve the DE $$yy^{\prime}+x=\sqrt{x^2+y^2},x>0$$ I show it is a homogenous first order DE and I use the substitution $y=Vx$. Then, I solve until $$\int \frac{V}{\sqrt{1+V^2}-(1+V^2)}dV=\int \frac{1}{x}dx$$ I don know how to integrate LHS. Any hint?

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Hint: Divide by $y$ your equation and then use $u=\frac{y}{x}$ – Tomas Jan 29 '13 at 9:43

Hint: Integral on left side is $\int \frac{zdz}{z-z^2}=\int\frac{dz}{1-z}=-\log|1-z|$ where $z^2=1+V^2$