Question

Solve the IVP (3x + 8)(y
2 + 4) dx − 4y(x
2 + 5x + 6) dy = 0, y(1) = 2.

Answer 1

Hi,

Explanation:

`(3x+8)(y^2+4)dx-4y(x^2+5x+6)dy=0\\\\4y(x^2+5x+6)dy=(3x+8)(y^2+4)dx\\\\(4y)/(y^2+4) dy=(3x+8)/((x+2)(x+3)) dx\\\\(3x+8)/((x+2)(x+3)) =(A)/(x+2)+(B)/(x+3) \\\\A=2\ and\ B=1\\\\2ln(y^2+4)=2ln(x+2)+ln(x+3)+C\\\\(y^2+4)^2=D*(x+2)^2(x+3)\\\\But\ y(1)=2\\\\2=\sqrt{√(D(x+2)^2(x+3))-4 } \\\\D=(16)/(9) \\\\\boxed{(y^2+4)^2=(16)/(9)*(x+2)^2(x+3)}\\`

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