y^'=15y-y^2-50

Lösung

y^{^{'}} = 15 y - y^{2} - 50

y = - \frac{10 e ^{5 t + 5 c_{1}}}{1 - e ^{5 (t + c_{1})}} + \frac{5}{1 - e ^{5 (t + c_{1})}}

Schritte zur Lösung

y^{'} (t) = 15 y - y^{2} - 50

Solve separable ODE:

y = - \frac{10 e ^{5 t + 5 c_{1}}}{1 - e ^{5 (t + c_{1})}} + \frac{5}{1 - e ^{5 (t + c_{1})}}

y = - \frac{10 e ^{5 t + 5 c_{1}}}{1 - e ^{5 (t + c_{1})}} + \frac{5}{1 - e ^{5 (t + c_{1})}}

\frac{x ^{2}}{y ^{2} - 3} \frac{d y}{d x} = \frac{1}{2 y}, y (1) = 2

x^{^{'}} = e^{t + x}

d y = x (2 y d x - x d y)

(x + e^{x}) y d x - d y = 0

\frac{df}{d x} = f (x)