Question

For the differential equation, find the general solution:

`dy/dx + y = 1(y != 1)`

Answer 1

`dy/dx + y = 1`

⇒ `dy/dx = - (y - 1)`

⇒ `dy/(y - 1) = - dx`                     ....(1)

Integrating (1) both sides, we get

⇒ `intdy/(y - 1) = - intdx`

⇒ log (y - 1) = -x + C₁

⇒ y - 1 = e^-x+C₁

⇒ y = 1 + e^-x . e^C1

Hence, y = 1 + Ce^-x, which is the required solution