y^'(x)=(ln(x))/(1+y^2),y(1)=0

解答

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

ln (x)

求解步骤

\frac{ln ( x )}{1 + y ^{2}} = \frac{ln ( x )}{1 + y ^{2}}

常数函数结果为常数

y (1) = 0

= \frac{ln ( x )}{1 + 0 ^{2}}

乘开

\frac{ln ( x )}{1 + 0 ^{2}} : ln (x)

= ln (x)

\frac{d x}{d t} = 5 - \frac{3 x}{1500}

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

y^{^{'}} = - \frac{x - 3}{y - 8}, y (3) = 10

+ x ye d y + 4 x d x = 0

y^{^{'}} = - x^{2} y^{2}