y^'y=y^2

Lösung

y^{^{'}} y = y^{2}

y = e^{t + c_{1}}

Schritte zur Lösung

y^{'} (t) y = y^{2}

Solve separable ODE:

y = e^{t + c_{1}}

y = e^{t + c_{1}}

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

(2 + cos (t)) (\frac{d y}{d t}) - y sin (t) = 0

a y^{^{'}} = - b y^{2} - c, y (0) = 5

\frac{d y}{d t} - y = 1, y = 0

d y - (1 + y) d x = 0