2sin(3x)sin(2y)dy=3cos(3x)cos(2y)dx

解答

2 sin (3 x) sin (2 y) d y = 3 cos (3 x) cos (2 y) d x

y = \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn, y = - \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn

求解步骤

2 sin (3 x) sin (2 y) d y = 3 cos (3 x) cos (2 y) d x

求解可分离方程:

y = \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn, y = - \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn

y = \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn, y = - \frac{arccos ( \frac{c _{1}}{s i n ( 3 x )} )}{2} + πn

7 x y \frac{d y}{d x} = 3, y > 0, x > 0

\frac{d x}{d t} + \frac{8 x}{240} = 2

\frac{d y}{d x} = 4 x - 7 x y

y^{^{'}} y = (x^{2} - 2 x + 2) (4 x^{3} - x^{2} + 5)

\frac{d y}{d x} = \frac{cos ( x )}{3 y}