(y^')/(y-y^2)= t/((1+t^2))

解答

\frac{y ^{^{'}}}{y - y ^{2}} = \frac{t}{( 1 + t ^{2} )}

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

求解步骤

\frac{y ^{'} ( t )}{y - y ^{2}} = \frac{t}{( 1 + t ^{2} )}

求解可分离方程:

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

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

t an g e n t f (x) = - 2 x cos (x), (π, 2 π)

n = 0 \sum \infty n! 2^{n}

5 x y (\frac{d y}{d x}) = 4

\frac{d}{d x} (5 e^{- 0.5 x} + 100)

\int \frac{2 ^{18 x}}{4} d x