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How to find the general solution of equation 3xy ^ 2dy = (y ^ 3-x ^ 2) DX?
Y = (- x ^ 2 + Cx) ^ (1 / 3), C is any constant
Solution steps:
3xy^2dy=(y^3-x^2)dx,
(3xy^2)*y'=y^3-x^2,
And [(y ^ 3) / x] '= [(3xy ^ 2) * y' - (y ^ 3)] / (x ^ 2) = - 1,
Let (y ^ 3) / x = - x + C, C be any constant,
That is y = (- x ^ 2 + Cx) ^ (1 / 3), C is any constant
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