(x^4+y^4)dx-(xy^3)dy=0

解答

(x^{4} + y^{4}) d x - (x y^{3}) d y = 0

y = x 4 4 ln (x) + c_{1}, y = - x 4 4 ln (x) + c_{1}

求解步骤

(x^{4} + y^{4}) d x - (x y^{3}) d y = 0

求解贝努力方程:

y = x 4 4 ln (x) + c_{1}, y = - x 4 4 ln (x) + c_{1}

y = x 4 4 ln (x) + c_{1}, y = - x 4 4 ln (x) + c_{1}

\int \frac{1}{( x ^{2} + 25 ) ^{2}} d x

\frac{\partial}{\partial y} (4 - x^{2} - 2 y^{2})

2 r (s^{2} + 1) d r + (r^{4} + 1) d s = 0

\int_{- \infty}^{15} x e^{\frac{x}{5}} d x

x \to 1 lim (\frac{1}{x ^{2} - x})