solve dy/dx = (y^2)(e^2y), y(0) = 1. e^2y = x lnx = 2y. lnx/2 = y dy/y^2 = (0.5)(lnx)dx -y^-1 = (.5)(xlnx-x) + C C = -1 y = -2/(xlnx -x -1) Is this correct? Im stuck here since lnx is not determinable.
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Originally Posted by lord12 solve dy/dx = (y^2)(e^2y), y(0) = 1. e^2y = x lnx = 2y. lnx/2 = y dy/y^2 = (0.5)(lnx)dx -y^-1 = (.5)(xlnx-x) + C C = -1 y = -2/(xlnx -x -1) Is this correct? Im stuck here since lnx is not determinable. $\displaystyle \frac{dy}{dx} = y^{2}e^{2y}, \ y(0) = 1 $ $\displaystyle \ln x = 2y \implies y = \frac{\ln x}{2} $. How did you get $\displaystyle \frac{dy}{y^2} = \frac{1}{2} \ln x \ dx $? According to your substitution, $\displaystyle \frac{dy}{y^2} = x \ dx $.
first, where does this come from? $\displaystyle x = e^{2y}$ or is it included in the given?
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