y^'-8y^3=0

Lösung

y^{^{'}} - 8 y^{3} = 0

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

Schritte zur Lösung

y^{'} (t) - 8 y^{3} = 0

Solve separable ODE:

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

y = - \frac{1}{2 ( 8 t + c _{1} )}, y = - - \frac{1}{2 ( 8 t + c _{1} )}

\int \frac{1}{2 x - 1 - 4 2 x - 1} d x

\int x^{1.1} (\frac{1}{3 x} - 1) d x

im pl i c i t x^{2} + y^{2} = 25

\int_{1}^{2} x (x - 1)^{9} d x

h \to 1 - lim (\frac{h - 2}{1 - h})