(3xcos(3x)+sin(3x)-3)dx+(2y+5)dy=0

Lösung

(3 x cos (3 x) + sin (3 x) - 3) d x + (2 y + 5) d y = 0

y = \frac{- 5 + 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}, y = \frac{- 5 - 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}

Schritte zur Lösung

(3 x cos (3 x) + sin (3 x) - 3) d x + (2 y + 5) d y = 0

Solve separable ODE:

y = \frac{- 5 + 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}, y = \frac{- 5 - 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}

y = \frac{- 5 + 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}, y = \frac{- 5 - 25 - 4 ( - c _{1} - 3 x + x sin ( 3 x ) )}{2}

sin (2 x) d y - e^{y} d x = 0

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

\frac{d x}{d y} = (6 x - 1) e^{y}

y^{^{'}} = 3 t 3 y^{2}

\frac{d y}{d x} = 3 x^{2} y^{2}, y! = 0