Question

For each of the differential equation, find the general solution: $\frac{dy}{dx}+ y =1(y\neq 1$

Answer 1

Answer:

$y = Ke^{-x} + 1$

Step-by-step explanation:

Hi,

Given differential equation is

$\frac{dy}{dx} + y = 1$ which can be written as

$\frac{dy}{dx} = 1 - y$

Separating out y terms to one side and x terms to other,

we get

$\frac{dy}{(1 - y)} = dx$

Integrating on  both sides, we get

$\int \frac{dy}{(1 - y)} = \int dx$

-㏑|y - 1| = x + c

or $y = Ke^{-x} + 1$

which is the required general solution.

Hope, it helps!