What is the general solution of the differential equation $dy/dx + 2y = 0$ ?

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Answer 1 · 2017-07-22T12:17:53

$y = Ce^(-2x)$

We have:

$dy/dx + 2y = 0$

We can just rearrange as follows:

$dy/dx = -2y => 1/y \ dy/dx = -2$

This is a first Order Separable Differential Equation and "separate the variables" to get

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

And integrating gives us:

$ln |y| = -2x + A$

$:. |y| = e^(-2x + A)$

Note that as $e^alpha gt 0 AA alpha in RR$ , we can write

$y = e^(-2x + A)$
$\ \ = e^(-2x)e^A$
$\ \ = Ce^(-2x)$

What is the general solution of the differential equation dy/dx + 2y = 0 dy/dx + 2y = 0?