How do you find the general solution to $\frac{d y}{d x} = \frac{2 x}{e^{2 y}}$ $dy/dx=(2x)/e^(2y)$ ?

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Answer 1 · 2016-07-24T17:38:05

$y = \frac{1}{2} ln (2 x^{2} + C)$ $y = 1/2 ln (2x^2 + C)$

$\frac{d y}{d x} = \frac{2 x}{e^{2 y}}$ $dy/dx=(2x)/e^(2y)$

this is separable

$e^{2 y} \frac{d y}{d x} = 2 x$ $e^(2y)dy/dx=2x$

$int \ e^(2y)dy/dx \ dx=int \ 2x \ dx$

$1/2 e^(2y) = x^2 + C$

$e^(2y) = 2x^2 + C$

$2y = ln (2x^2 + C)$

$y = 1/2 ln (2x^2 + C)$

How do you find the general solution to dy/dx=(2x)/e^(2y)dydx=2xe2ydy/dx=(2x)/e^(2y)?