3y^{''}-24y^'+48y=0

Lösung

3 y^{^{''}} - 24 y^{^{'}} + 48 y = 0

y = c_{1} e^{4 t} + c_{2} t e^{4 t}

Schritte zur Lösung

3 y^{''} (t) - 24 y^{'} (t) + 48 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} t e^{4 t}

y = c_{1} e^{4 t} + c_{2} t e^{4 t}

4 \frac{d ^{2} y}{d t ^{2}} - y = 0

y^{^{'''}} - 20 y^{^{''}} + 111 y^{^{'}} - 180 y = 0

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

\frac{d ^{4} y}{d x ^{4}} + 2 y^{^{'''}} - 2 y^{^{'}} - y = 0

y^{^{''}} + 6 y^{^{'}} + 7 y = 0