Question

Particular integral if dy/dx + y =1 is

Answer 1

In question

 \(\frac{dy}{dx}+y =1\)

⇒ \(∫\frac{dy}{1-y}=∫dx\)

⇒ -log |1-y| = x + log C

⇒ x = log c - log |1-y| = log\((\frac{c}{1-y})\)

⇒ \(\frac{c}{1-y}\) e^x

⇒ 1 - y = ce^-x 

⇒ y = 1 - ce^-x