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JцstinBieber
[tex]\displaystyle\frac{dy}{dx} = \frac{\ln x}{xy}, y(1)= 2.\ \int y\, dy = \int \frac{\ln x}{x} \, dx \ \Rightarrow\ \textstyle \frac{1}{2}y^2 = \frac{1}{2}( \ln x )^2 + C. \\ \\ \text{Now }y(1) = 2\ \Rightarrow \frac{1}{2}(2)^2 = \frac{1}{2}( \ln 1 )^2 + C\ \Righarrow\ 2 = C,\text{ so } \frac{1}{2}y^2 = \frac{1}{2} ( \ln x )^2 + 2. \\ \\ y^2 = ( \ln x )^2 + 4\ \Rightarrow\ y = \pm\sqrt{ (\ln x)^2 + 4}. \\ \\ \text{ Since $y(1) = 2$, we have } y = \sqrt{ (\ln x)^2 + 4}[/tex]

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