(y-1)y^'=1

Lösung

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

y = 1 + 2 (t + c_{1}) + 1, y = 1 - 2 (t + c_{1}) + 1

Schritte zur Lösung

(y - 1) y^{'} (t) = 1

Solve separable ODE:

y = 1 + 2 (t + c_{1}) + 1, y = 1 - 2 (t + c_{1}) + 1

y = 1 + 2 (t + c_{1}) + 1, y = 1 - 2 (t + c_{1}) + 1

\frac{d N}{d t} = 0.005 N

\frac{d y}{d x} = y^{2} x cos (x^{2})

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

2 y^{^{'}} + y = 5, y (3) = 0

x y^{2} \frac{d y}{d x} = x + 6