Differential Equations

It is given that
y′(x) + y(x)g′(x) = g(x)g′(x)
⇒ eg(x) y' (x) + eg(x) g' (x) y (x) = eg(x) g (x) g' (x)
⇒

d

dx

(y(x)eg(x)) = eg(x) g (x) g' (x)
Therefore, y (x) = eg(x) = ∫eg(x)g(x)g′(x)dx
= ∫etdt, [where g (x) = t]
Therefore, y(x)eg(x) = (g(x)−1)eg(x)+c
Substituting x = 0⇒0 = (0 −1) × 1 + c ⇒ c = 1.
Substituting x = 2 ⇒ y(2) ×1 = (0 −1) × (1) +1.
Hence, y(2) = 0.