y^'= 1/(1-3y)

Lösung

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

y = - \frac{- 1 + - 6 ( t + c _{1} ) + 1}{3}, y = \frac{- 6 ( t + c _{1} ) + 1 + 1}{3}

Schritte zur Lösung

y^{'} (t) = \frac{1}{1 - 3 y}

Solve separable ODE:

y = - \frac{- 1 + - 6 ( t + c _{1} ) + 1}{3}, y = \frac{- 6 ( t + c _{1} ) + 1 + 1}{3}

y = - \frac{- 1 + - 6 ( t + c _{1} ) + 1}{3}, y = \frac{- 6 ( t + c _{1} ) + 1 + 1}{3}

\frac{d y}{y} = 35 x^{4} d x

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

x^{^{'}} + sin (t) \cdot x - sin (t) = 0

x^{2} d x - (y + x^{3} y) d y = 0

\frac{d x}{d t} = (1 + t) (1 + x)