Solve the initial value problem. x squared StartFraction dy Over dx EndFraction equalsStartFraction 4 x squared minus x minus 3 Over (x plus 1) (y plus 1) EndFraction , y (1) equals 2 The solution is nothing. (Type an implicit solution. Type an equation using x and y as the variables.)

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