y^'=12y-15

Lösung

y^{^{'}} = 12 y - 15

y = \frac{e ^{12 t + c_{1}}}{4} + \frac{5}{4}

Schritte zur Lösung

y^{'} (t) = 12 y - 15

Solve separable ODE:

y = \frac{e ^{12 t + c_{1}}}{4} + \frac{5}{4}

y = \frac{e ^{12 t + c_{1}}}{4} + \frac{5}{4}

(x - 2) y^{^{'}} - y - 1 = 0

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

y^{^{'}} = x y + \frac{4 x}{y}, y (0) = 1

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

(x^{2} + 1) y + 2 x y^{^{'}} + (e^{x}) x y = 0