How do you find the general solution to $dy/dx=2y-1$ ?

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Answer 1 · 2016-08-01T21:56:26

$y = Ce^(2x) + 1/2$

$dy/dx=2y-1$

separating the variables
$1/(2y-1) *dy/dx=1$

integrating
$int \ 1/(2y-1) dy/dx \ dx =int \ dx$

$int \ 1/(2y-1) \ dy =int \ dx$

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

$ln(2y-1) =2x + C$

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

$y-1/2 = Ce^(2x)$

$y = Ce^(2x) + 1/2$

How do you find the general solution to dy/dx=2y-1dy/dx=2y-1?