y^5+9y^'=0

Lösung

y^{5} + 9 y^{^{'}} = 0

y = 4 - \frac{9}{4 ( - t + c _{1} )}, y = - 4 - \frac{9}{4 ( - t + c _{1} )}

Schritte zur Lösung

y^{5} + 9 y^{'} (t) = 0

Solve separable ODE:

y = 4 - \frac{9}{4 ( - t + c _{1} )}, y = - 4 - \frac{9}{4 ( - t + c _{1} )}

y = 4 - \frac{9}{4 ( - t + c _{1} )}, y = - 4 - \frac{9}{4 ( - t + c _{1} )}

\frac{d y}{d x} = 4 e^{x - y}, y (0) = ln (6)

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

y^{^{'}} = \frac{3 y}{x + 6}

x^{^{'}} = - x + 1

4 y x \frac{d x}{d y} = x^{2} + 1