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How general solution of this form of differential equation can be found? $$\frac {dy}{dx} = \frac {2y}{x}$$

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$$\frac{dy}{dx}=\frac{2y}{x} \hspace{10pt}\Leftrightarrow \hspace{10pt}\frac{dy}{2y}=\frac{dx}{x}$$ Now you just need to integrate.

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@TPStar: It is a separable ODE. –  B. S. Nov 23 '12 at 17:08
    
Indeed, that's what I wrote... –  Dennis Gulko Nov 23 '12 at 17:18
    
okay that becomes $$\frac {y^2}{4}dy = \log xdx$$ am i right? –  TPSstar Nov 23 '12 at 17:28
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No, That becomes $\frac12\ln y=\ln x+C$ and so $y=Cx^2$ –  Dennis Gulko Nov 23 '12 at 17:32
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