(xy^2-x)dx+(2x^2+4)dy=0

解答

(x y^{2} - x) d x + (2 x^{2} + 4) d y = 0

y = \frac{e ^{c_{1}} ( ( x ^{2} + 2 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} + 2 - e ^{2 c_{1}}} + \frac{x ^{2} + 2 + e ^{c_{1}} ( x ^{2} + 2 ) ^{\frac{1}{2}}}{x ^{2} + 2 - e ^{2 c_{1}}}

求解步骤

(x y^{2} - x) d x + (2 x^{2} + 4) d y = 0

求解可分离方程:

y = \frac{e ^{c_{1}} ( ( x ^{2} + 2 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} + 2 - e ^{2 c_{1}}} + \frac{x ^{2} + 2 + e ^{c_{1}} ( x ^{2} + 2 ) ^{\frac{1}{2}}}{x ^{2} + 2 - e ^{2 c_{1}}}

y = \frac{e ^{c_{1}} ( ( x ^{2} + 2 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} + 2 - e ^{2 c_{1}}} + \frac{x ^{2} + 2 + e ^{c_{1}} ( x ^{2} + 2 ) ^{\frac{1}{2}}}{x ^{2} + 2 - e ^{2 c_{1}}}

\frac{d y}{d x} = \frac{3 x ^{2} - e ^{x}}{2 y - 5}

(x^{2} y + 2 y) y^{^{'}} = 2 x y^{2} + 2 x

\frac{d y}{d x} = e^{x + y} + \frac{e ^{y} x}{1 + x ^{2}}

sec^{2} (x) y^{^{'}} = - csc (y)

y^{^{'}} = e^{x - y}, y (0) = 1