cos(3y)dy+sin(2x)dx=0

Lösung

cos (3 y) d y + sin (2 x) d x = 0

y = \frac{arcsin ( \frac{3}{2} cos ( 2 x ) + c _{1} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) - c _{1} )}{3} + \frac{2 πn}{3}

Schritte zur Lösung

cos (3 y) d y + sin (2 x) d x = 0

Solve separable ODE:

y = \frac{arcsin ( \frac{3}{2} cos ( 2 x ) + c _{1} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) - c _{1} )}{3} + \frac{2 πn}{3}

y = \frac{arcsin ( \frac{3}{2} cos ( 2 x ) + c _{1} )}{3} + \frac{2 πn}{3}, y = \frac{π}{3} + \frac{arcsin ( - \frac{3}{2} cos ( 2 x ) - c _{1} )}{3} + \frac{2 πn}{3}

\frac{d x}{d t} + cos (t) x = cos (t), x (0) = 2

\frac{d r}{d t} = - 4 r t, r = r_{0} (t), t = 0

y^{^{'}} - \frac{x ^{2} y - y}{y + 4} = 0, y (3) = 1

(e^{m} + 1)^{2} e^{- m} d t + (e^{t} + 1)^{3} e^{- t} d m = 0

\frac{d y}{d x} = \frac{1 + x}{x y ^{5}}