(d^2+5d+6)y=(d+1)6t^2

Lösung

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{19}{18} - \frac{5}{3} t - \frac{5}{3} d t + \frac{19}{18} d + d t^{2} + t^{2}

Schritte zur Lösung

(d^{2} + 5 d + 6) y = (d + 1) 6 t^{2}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{19}{18} - \frac{5}{3} t - \frac{5}{3} d t + \frac{19}{18} d + d t^{2} + t^{2}

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{19}{18} - \frac{5}{3} t - \frac{5}{3} d t + \frac{19}{18} d + d t^{2} + t^{2}

y^{^{''}} + y = 8 e^{- 6 x}

(d^{2} - 1) \cdot y = x \cdot e^{x} + cos^{2} (x) + x^{2}

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = x^{2} + 1

y^{^{''}} - y^{^{'}} = \frac{1}{e ^{x} + 1}