$y=2\sin\sqrt{x}$
$y'=2\cos\sqrt{x}.(\sqrt{x})'$
$=2\cos\sqrt{x}.\dfrac{1}{2\sqrt{x}}$
$=\dfrac{\cos\sqrt{x}}{\sqrt{x}}$
$4x(y')^2+y^2=4x.\dfrac{\cos^2\sqrt{x} }{x}+4\sin^2\sqrt{x}$
$=4\cos^2\sqrt{x}+4\sin^2\sqrt{x}$
$=4(\sin^2\sqrt{x}+\cos^2\sqrt{x})$
$=4$
Vậy $4x(y')^2+y^2=4$