Solve the Differential Equation $dy/dx = 2xy - x$ ?

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Answer 1 · 2017-12-21T12:44:51

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

We have:

$dy/dx = 2xy - x$

We can factorize the RHS to get

$dy/dx = x(2y - 1)$

Which is separable, so we can "separate the variables" to get:

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

So we integrate

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

And we can rearrange:

$ln(2y-1) = x^2 + 2C$
$:. 2y-1 = e^(x^2 + 2C)$
$:. 2y-1 = e^(x^2)e^(2C)$
$:. y = Ae^(x^2) + 1/2$

Solve the Differential Equation dy/dx = 2xy - x dy/dx = 2xy - x ?