(dx)/(dt)+4tx^3+x/t =0

解答

\frac{d x}{d t} + 4 t x^{3} + \frac{x}{t} = 0

x = \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}, x = - \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}

求解步骤

\frac{d x}{d t} + 4 t x^{3} + \frac{x}{t} = 0

求解贝努力方程:

x = \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}, x = - \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}

x = \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}, x = - \frac{1}{t ^{2} ( 8 ln ( t ) + c _{1} )}

f (x) = \frac{1 - sec ( x )}{tan ( x )}

d er i v a t i v e f (x) = 7 + \frac{2}{x} + \frac{5}{x ^{2}}

\int \frac{1}{( x - 9 ) ( 3 - x )} d x

x y^{^{''}} + 2 y^{^{'}} = 24 x^{2}

\int - 5 x d x