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Question
For the differential equation, find the general solution:
`dy/dx + y = 1(y != 1)`
Solution
`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 . eC1
Hence, y = 1 + Ce-x, which is the required solution
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