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) #