(3x^2-2xy)dx+(2y-x^2)dy=0

解答

(3 x^{2} - 2 x y) d x + (2 y - x^{2}) d y = 0

y = \frac{x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

求解步骤

(3 x^{2} - 2 x y) d x + (2 y - x^{2}) d y = 0

求解恰当方程:

y = \frac{x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

y = \frac{x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

\int_{- 1}^{3} (x^{2} - 2 x) d x

\int_{0}^{3} \frac{9}{x ^{2} + 9} d x

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y^{^{'}} = \frac{18 ( 1 + 3 y )}{( 3 - 3 x ) ^{2}}, y (2) = 0