The correct answer is: \(ye^{2\sqrt{x}}=2\sqrt{x}+C\) \([\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}]\frac{dx}{dy}=1\) \(\implies \frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\) \(⇒\frac{dy}{dx}+\frac{y}{\sqrt{x}}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\) This equation is a linear differential equation of the form \(\frac{dy}{dx}+py=Q\),where \(p=\frac{1}{\sqrt{x}}\) and \(Q=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.\) Now,\(I.F.=e^{∫pdx}=e^{∫\frac{1}{\sqrt{x}}dx}=e^{2\sqrt{x}}\) The general solution of the given differential equation is given by, \(y(I.F.)=∫(Q×I.F.)dx+C\) \(⇒ye^{2\sqrt{x}}=∫(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}×e^{2\sqrt{x}})dx+C\) \(⇒ye^{2\sqrt{x}}=∫\frac{1}{\sqrt{x}}dx+C\) \(⇒ye^{2\sqrt{x}}=2\sqrt{x}+C\)