Question

Solve the differential equation `dy/dx + y = 0` ?

Answer 1

`y = Ae^(-x)`

Explanation:

We can rewrite the equation:

`dy/dx + y = 0`

as:

`dy/dx = -y => 1/ydy/dx=-1``

Which is a First Order linear separable Differential Equation, so we can "separate the variables" to get:

`int \ 1/y \ = int \ -1 \ dx`

Which we can integrate to get:

`ln |y| = -x + c`

Taking Natural logarithms we then get:

`|y| = e^(-x + c)`

As `e^x > 0 AA x in RR` we then get:

`y = e^(-x + c)`
`\ \ = Ae^(-x)`

We can easily verify the solution:

`y = Ae^(-x) => y' = -Ae^(-x)`
`y' + y = -Ae^(-x) + Ae^(-x) = 0 \ \ \ QED`

Answer 2

`y = Ce^-x; C > 0`

Explanation:

Given: `dy/dx + y = 0`

Subtract y from both sides of the equation:

`dy/dx = -y`

We shall use the separation of variables method.

To do this, we multiply both sides by `dx` and divide both sides by `y`:

`dy/y = -dx`

Please observe that we have "separated" all of the things to with y on the left and all of the things to do with x on the right.

We can integrate both sides:

`intdy/y = -intdx`

The left side becomes the natural logarithm and the right becomes x plus an arbitrary constant:

`ln|y| = -x+c`

Eliminate the logarithm by making both sides exponents of e:

`e^(ln|y|) = e^(-x+c)`

By the definition of the inverse function of a function the left side becomes y:

`y = e^(-x+c)`

We can write the some of two exponents as a product:

`y = (e^c)(e^-x)`

e to an arbitrary constant is just another constant but is can only be positive, because that is the range of e:

`y = Ce^-x; C > 0`

Check the solution:

`dy/dx = -Ce^-x`

Substitute y and `dy/dx` into the original equation:

`-Ce^-x + Ce^-x= 0`

`0= 0`

This checks.