`dy/dx = (x-3) / (y(y+4))` Solve the first-order differential equation by any appropriate method

Given ,

`dy/dx = (x-3) / (y(y+4))`

=> `y(y+4) dy = (x-3)dx`

on integrating on both sides we get

=> `int y(y+4) dy =int (x-3)dx`

=> `int ( y^2+4y )dy=int (x-3)dx`

=> `y^3/3 +4y^2/2+c_1=x^2/2 -3x+c_2`

=>` y^3 /3 +2y^2+c_1 = x^2/2 -3x+c_2`

=>` y^3 /3 +2y^2+c_1 - x^2/2 +3x-c_2=0`

=>` y^3 /3 +2y^2+c_1 = x^2/2 - 3x+c_2`

=>` y^3 /3 +2y^2= x^2/2 - 3x+C` where `C = c_2-c_1` is an arbitrary constant.

=>` y^3 +6y^2 = 3x^2/2 - 9x+C` is the final answer

Comments

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