y^3-y^'=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d y}{d x} = y \cdot x^{2} - 1.1 \cdot y

(2 x y + 2 x) d x = y d y

y^{2} + y = t \cdot y^{^{'}}, y (1) = - 2

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

\frac{d x}{d t} = 1.8 - \frac{5 x}{100}