Question

help me with the following equation: xyy'+√(1+y^2 )=0 solve for y

Answer 1

\[xyy'+\sqrt{1+y^2}=0\]\[x\frac{yy'}{\sqrt{1+y^2}}+1=0\]\[x\frac{(y^2)'}{2\sqrt{1+y^2}}+1=0\]\[\frac{(y^2)'}{2\sqrt{1+y^2}}=-\frac{1}{x}\]\[\left(\sqrt{1+y^2}\right)'=-\frac{1}{x}\]\[\sqrt{1+y^2}=-\ln{x}+C=\ln{\frac{C_1}{x}}\]