(1+x^2)y^'+4xy=(1+x^2)^{-2}

解答

(1 + x^{2}) y^{^{'}} + 4 x y = (1 + x^{2})^{- 2}

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

求解步骤

(1 + x^{2}) y^{'} (x) + 4 x y = (1 + x^{2})^{- 2}

求解线性方程:

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

\frac{d}{d x} (e^{2 x^{2} - 1})

d er i v a t i v e f (t) = 3 t

\int_{0}^{8} \frac{1}{64 + x ^{2}} d x

\frac{d}{d x} (\frac{x sin ( 2 x )}{cos ( 3 x )})

8 t \frac{d y}{d t} + y = t^{5}