y^'=(e^{5x})/(cos(2y))

解答

y^{^{'}} = \frac{e ^{5 x}}{cos ( 2 y )}

y = \frac{arcsin ( \frac{2}{5} e ^{5 x} + c _{1} )}{2} + πn, y = \frac{π}{2} + \frac{arcsin ( - \frac{2}{5} e ^{5 x} - c _{1} )}{2} + πn

求解步骤

y^{'} (x) = \frac{e ^{5 x}}{cos ( 2 y )}

求解可分离方程:

y = \frac{arcsin ( \frac{2}{5} e ^{5 x} + c _{1} )}{2} + πn, y = \frac{π}{2} + \frac{arcsin ( - \frac{2}{5} e ^{5 x} - c _{1} )}{2} + πn

y = \frac{arcsin ( \frac{2}{5} e ^{5 x} + c _{1} )}{2} + πn, y = \frac{π}{2} + \frac{arcsin ( - \frac{2}{5} e ^{5 x} - c _{1} )}{2} + πn

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

y^{^{'}} - 3 y = 2

y + y^{^{'}} - 2 y = 0

(2 x y) d x = (3 - x^{2}) d y

y^{^{'}} = \frac{2 x}{y ^{2}}