y^{''}+10y^'+25y=0,y(-1)=6,y^'(-1)=5

Lösung

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 5

y = 41 e^{- 5 t - 5} + 35 e^{- 5 t - 5} t

Schritte zur Lösung

y^{''} (t) + 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = 41 e^{- 5 t - 5} + 35 e^{- 5 t - 5} t

y = 41 e^{- 5 t - 5} + 35 e^{- 5 t - 5} t

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 6, y^{^{'}} (- 1) = 4

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

\frac{d ^{4} y}{d x ^{4}} - a \frac{d ^{2} y}{d x ^{2}} - b y = 0

x^{^{''}} + 2 x^{^{'}} + 0.75 x = 0

y^{^{''}} = 7 y^{^{'}} - 12 y, y (0) = 0, y^{^{'}} (0) = - 1