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

May 14, 2017

$y = A {e}^{- x}$

#### Explanation:

We can rewrite the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$

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:

$\int \setminus \frac{1}{y} \setminus = \int \setminus - 1 \setminus \mathrm{dx}$

Which we can integrate to get:

$\ln | y | = - x + c$

Taking Natural logarithms we then get:

$| y | = {e}^{- x + c}$

As ${e}^{x} > 0 \forall x \in \mathbb{R}$ we then get:

$y = {e}^{- x + c}$
$\setminus \setminus = A {e}^{- x}$

We can easily verify the solution:

$y = A {e}^{- x} \implies y ' = - A {e}^{- x}$
$y ' + y = - A {e}^{- x} + A {e}^{- x} = 0 \setminus \setminus \setminus Q E D$

May 14, 2017

y = Ce^-x; C > 0

#### Explanation:

Given: $\frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$

Subtract y from both sides of the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - y$

We shall use the separation of variables method.

To do this, we multiply both sides by $\mathrm{dx}$ and divide both sides by $y$:

$\frac{\mathrm{dy}}{y} = - \mathrm{dx}$

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:

$\int \frac{\mathrm{dy}}{y} = - \int \mathrm{dx}$

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:

$y = \left({e}^{c}\right) \left({e}^{-} x\right)$

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:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - C {e}^{-} x$

Substitute y and $\frac{\mathrm{dy}}{\mathrm{dx}}$ into the original equation:

$- C {e}^{-} x + C {e}^{-} x = 0$

$0 = 0$

This checks.