1/((1+2y^2))y^'=csc(x)

解答

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

y = \frac{tan ( 2 ( ln ( tan ( \frac{x}{2} ) ) + c _{1} ) )}{2}

求解步骤

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

求解可分离方程:

y = \frac{tan ( 2 ( ln ( tan ( \frac{x}{2} ) ) + c _{1} ) )}{2}

y = \frac{tan ( 2 ( ln ( tan ( \frac{x}{2} ) ) + c _{1} ) )}{2}

x \to 0 lim ((x + e^{\frac{x}{6}})^{\frac{6}{x}})

\int (a + b x^{3})^{3} d x

x \to 3 lim (- 2 x^{3})

d er i v a t i v e \frac{1}{x + 4}

\int \frac{20}{( x - 1 ) ( x ^{2} + 9 )} d x