Answer:
[tex]y(x) = e^{-x} +C[/tex]
Step-by-step explanation:
As per the question,
Given first-order differential equation is
y' = -y
That is,
[tex]\frac{dy}{dx} + y = 0[/tex]
[tex]\frac{dy}{y} = -dx[/tex]
On Integrating both side,
[tex]\int \frac{dy}{y} = \int -dx[/tex]
As we know that,
[tex]\int \frac{1}{x} dx = ln\ x +C[/tex]
Therefore,
[tex]\int \frac{dy}{y} = \int -dx[/tex]
[tex]ln\ y = e^{-x} + C[/tex]
Hence the required function is
[tex]y(x) = e^{-x} +C[/tex]
