y^'=y^2+y-30

Lösung

y^{^{'}} = y^{2} + y - 30

y = - \frac{5 e ^{- 11 t - 11 c_{1}}}{1 - e ^{- 11 (t + c_{1})}} - \frac{6}{1 - e ^{- 11 (t + c_{1})}}

Schritte zur Lösung

y^{'} (t) = y^{2} + y - 30

Solve separable ODE:

y = - \frac{5 e ^{- 11 t - 11 c_{1}}}{1 - e ^{- 11 (t + c_{1})}} - \frac{6}{1 - e ^{- 11 (t + c_{1})}}

y = - \frac{5 e ^{- 11 t - 11 c_{1}}}{1 - e ^{- 11 (t + c_{1})}} - \frac{6}{1 - e ^{- 11 (t + c_{1})}}

x \frac{d y}{d x} + 4 y = 0

y^{^{'}} = x y^{3} (1 + x^{2})

\frac{d y}{y} = \frac{d x}{1 + x}

\frac{d y}{d x} y = e^{x^{2} l n (x)}

y^{^{'}} = - x y + 4 x