(1+x^4)dy+(1+4y^2)dx=0

解答

(1 + x^{4}) d y + (1 + 4 y^{2}) d x = 0

y = \frac{tan ( - c _{1} + \frac{l n ( \frac{x ^{2} - 2 x + 1}{x ^{2} + 2 x + 1} ) - 2 a r c t a n ( 2 x + 1 ) - 2 a r c t a n ( 2 x - 1 )}{2 2} )}{2}

求解步骤

(1 + x^{4}) d y + (1 + 4 y^{2}) d x = 0

求解可分离方程:

y = \frac{tan ( - c _{1} + \frac{l n ( \frac{x ^{2} - 2 x + 1}{x ^{2} + 2 x + 1} ) - 2 a r c t a n ( 2 x + 1 ) - 2 a r c t a n ( 2 x - 1 )}{2 2} )}{2}

y = \frac{tan ( - c _{1} + \frac{l n ( \frac{x ^{2} - 2 x + 1}{x ^{2} + 2 x + 1} ) - 2 a r c t a n ( 2 x + 1 ) - 2 a r c t a n ( 2 x - 1 )}{2 2} )}{2}

y^{^{'}} - 5 y^{(2)} e^{(- 3 x)} = 0, y (0) = 9

\int \frac{x ^{3}}{x ^{2} + 2 x + 1} d x

t an g e n t y = \frac{( x ^{2} )}{8 + x}

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

\int a e^{- m x} d x