y^'=-2/4 x^3y^8

解答

y^{^{'}} = - \frac{2}{4} x^{3} y^{8}

y = - 7 \frac{8}{7 ( - x ^{4} + c _{1} )}

求解步骤

y^{'} (x) = - \frac{2}{4} x^{3} y^{8}

求解可分离方程:

y = - 7 \frac{8}{7 ( - x ^{4} + c _{1} )}

y = - 7 \frac{8}{7 ( - x ^{4} + c _{1} )}

y^{^{'}} = - t y^{3}

\frac{d y}{d x} y = (3 + x)^{\frac{5}{x}}

f^{^{'}} (x) 4 f (x) = 2 x - 6

y y^{^{'}} = (1 - y^{2}) sin (x)

H (x) x d x - x (- ln (H (x))) (H (x) d x + x d H (x)) = 0