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

Lösung

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

y = 1 + c_{1} x^{2} + 1, y = 1 - c_{1} x^{2} + 1

Schritte zur Lösung

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

Solve separable ODE:

y = 1 + c_{1} x^{2} + 1, y = 1 - c_{1} x^{2} + 1

y = 1 + c_{1} x^{2} + 1, y = 1 - c_{1} x^{2} + 1

x \cdot \frac{d x}{d t} = e^{3 x} t

x^{^{'}} = e^{x} \cdot t \cdot cos (t)

\frac{d P}{d t} = 4 Pt, P (1) = 6

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

2 (\frac{d q}{d t}) + 6 q = 2