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27 August, 09:17

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.)

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  1. 27 August, 12:28
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    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
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