y^'=(2y)/(t-1)

Lösung

y^{^{'}} = \frac{2 y}{t - 1}

y = e^{c_{1}} t^{2} - 2 e^{c_{1}} t + e^{c_{1}}

Schritte zur Lösung

y^{'} (t) = \frac{2 y}{t - 1}

Solve separable ODE:

y = e^{c_{1}} t^{2} - 2 e^{c_{1}} t + e^{c_{1}}

y = e^{c_{1}} t^{2} - 2 e^{c_{1}} t + e^{c_{1}}

y^{^{'}} = 5 t^{3} y^{\frac{1}{5}}, y (0) = 0

\frac{d y}{d x} = \frac{- 4 y + 9}{5 x + 4}

2 y + 3 + (2 x - 5) \frac{d y}{d x} = 0

\frac{d y}{d x} + \frac{y}{x} = 0

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