[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}

y^{^{'}} = \frac{1 - y ^{2}}{1 - x ^{2}}

x y^{2} y^{^{'}} = x + x^{3}, y (0) = 1

y^{^{'}} = \frac{y - 1}{x + 3}

y^{^{'}} = 5 e^{- x - y}

(t^{2} - 22 t + 85) \frac{d y}{d t} = y