devarshi-dt-logo

Question:

Let y(x) be a solution of the differential equation (1+e^x)y' + ye^x = 1. If y(0) = 2, then which of the following statements is (are) true?

y(-∞)=0

y(∞)=0

y(x) has a critical point in the interval (-∞,0)

y(x) has no critical point in the interval (-∞,0)

Solution:

(1+e^x)dy/dx + ye^x = 1
⇒ dy/dx + e^x/(1+e^x)y = 1/(1+e^x)
I.F = e∫e^x/(1+e^x)dx = e^(ln(1+e^x)) = 1+e^x
Thus solution is, y(1+e^x) = ∫dx + c = x + c
Given y(0) = 2 ⇒ c = 2 ⇒ y = (x+2)/(1+e^x)
Clearly y(∞) = 0
Also dy/dx = (1+e^x) - e^x(x+2)/(1+e^x)^2
For critical point dy/dx = 0 ⇒ (1+e^x) - e^x(x+2) = 0
Let g(x) = (1+e^x) - e^x(x+2) ∴ g(-∞) = 1 - 0 > 0 and g(0) = 2 - 2 = 0 < 0
Hence y(x) possess a critical point in the interval (-∞,0)