y^'+6y^3=0

Lösung

y^{^{'}} + 6 y^{3} = 0

y = - \frac{1}{2 ( - 6 t + c _{1} )}, y = - - \frac{1}{2 ( - 6 t + c _{1} )}

Schritte zur Lösung

y^{'} (t) + 6 y^{3} = 0

Solve separable ODE:

y = - \frac{1}{2 ( - 6 t + c _{1} )}, y = - - \frac{1}{2 ( - 6 t + c _{1} )}

y = - \frac{1}{2 ( - 6 t + c _{1} )}, y = - - \frac{1}{2 ( - 6 t + c _{1} )}

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

\frac{d v}{d t} = 0.075 v - 9000

2 x d x + 2 y d y = 0

x^{2} (x d y + y d x) = 9 x d y - 9 y d x

\frac{d y}{d x} + 8 y = 0