(x^2-1)y^'=y

Lösung

(x^{2} - 1) y^{^{'}} = y

y = \frac{c _{1} x - 1}{x + 1}

Schritte zur Lösung

(x^{2} - 1) y^{'} (x) = y

Solve separable ODE:

y = \frac{c _{1} x - 1}{x + 1}

y = \frac{c _{1} x - 1}{x + 1}

\frac{d y}{d x} y = u, u = x^{2} + 10

y^{^{'}} + 5 y + 4 y = 0

\frac{d y}{d x} + \frac{y}{50} = 4, y (0) = 30

\frac{d V}{d t} = 80 - \frac{8}{100} V

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