Answer:
C = 2*Ln (2) - 1
Step-by-step explanation:
Given
x²(dy/dx) = (4x²-x-3)/(x+1)(y+1)
y(1) = 2
We apply separation of variables as follows
(y+1) dy = ((4x²-x-3)/(x+1)(x²)) dx
⇒ ∫(y+1) dy = ∫((4x²-x-3)/(x+1)(x²)) dx
(y²/2) + y + C₁ = 2 ∫(1/(x+1)) dx + ∫((2x-3)/x²) dx
⇒ (y²/2) + y + C₁ = 2 Ln (x+1) 2 Ln (x) + (3/x) + C₂
⇒ C₁ - C₂= Ln (x+1)² + Ln (x)² + (3/x) - (y²/2) - y
⇒ C = Ln ((x+1)²(x)²) + (3/x) - (y²/2) - y
⇒ C = Ln ((x²+x)²) + (3/x) - (y²/2) - y
if y(1) = 2
we get
C = Ln ((1²+1)²) + (3/1) - (2²/2) - 2
⇒ C = 2*Ln (2) + 3 - 4 = 2*Ln (2) - 1
