y^{''}+4y^'+4y=0,y(-1)=3,y^'(-2)=5

Lösung

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5

y = (3 + \frac{5 + 6 e ^{2}}{3 e ^{2}}) e^{- 2 t - 2} + \frac{( 5 + 6 e ^{2} ) e ^{- 2 t - 4} t}{3}

Schritte zur Lösung

y^{''} (t) + 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (3 + \frac{5 + 6 e ^{2}}{3 e ^{2}}) e^{- 2 t - 2} + \frac{( 5 + 6 e ^{2} ) e ^{- 2 t - 4} t}{3}

y = (3 + \frac{5 + 6 e ^{2}}{3 e ^{2}}) e^{- 2 t - 2} + \frac{( 5 + 6 e ^{2} ) e ^{- 2 t - 4} t}{3}

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

6 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 0

y^{^{''}} - 4 y^{^{'}} - a^{2} y = 0

0.5 y^{^{''}} + 2 y^{^{'}} + 6 y = 0

y^{^{'''}} - 2 y^{^{''}} - 15 y^{^{'}} = 0