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OpenStudy (anonymous):
ydx + (2xy - e^(-2y))dy = 0 see comment below for what I have done so far.
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OpenStudy (anonymous):
\[_{My} = 1 and {Nx} = 2y \] then I took (2y-1)/y = dmu = ln|mu| = 2y -ln|y| Is the integrating factor of \[\mu = e^{2y} - y\] correct? When I try to multiply it through the equation I still am not getting an exact equation.
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