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

Answer 1

Ezra Adams

# 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) #

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Answer 2

Samantha Adkison

The general solution of the differential equation ( \frac{dy}{dx} + 2y = 0 ) is ( y(x) = Ce^{-2x} ), where ( C ) is an arbitrary constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.