xy^3+(1-x)y^'=0

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{2 y}{x - 2}

\frac{d}{d θ} (r (θ)) = \frac{r ( θ ) ^{2}}{θ}, r (1) = 2

x f^{^{'}} (x) = (f (x))^{2}

\frac{d y}{d x} = 8 x y - 12 x^{2} y

x^{2} + 1 d x - 3 y d y = 0