Question

A solution of y' = -y is the function y(x)= . . .

Answer 1

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

Explanation:

As per the question,

Given first-order differential equation is

y' = -y

That is,

`(dy)/(dx) + y = 0`

`(dy)/(y) = -dx`

On Integrating both side,

`\int (dy)/(y) = \int -dx`

As we know that,

`\int (1)/(x) dx = ln\ x +C`

Therefore,

`\int (dy)/(y) = \int -dx`

`ln\ y = e^(-x) + C`

Hence the required function is

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