y^'=(3x^2+2x-4)/(2y-4),y(2)=0

解答

y^{^{'}} = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}, y (2) = 0

y = 2 - x (x^{2} + x - 4)

求解步骤

y^{'} (x) = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}

求解可分离方程:

y = 2 - x (x^{2} + x - 4)

y = 2 - x (x^{2} + x - 4)

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