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

Lösung

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

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

Schritte zur Lösung

4 y^{''} (t) - 5 y^{'} (t) + y = 0

Solve linear ODE:

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

y = (4 - c_{2}) e^{t} + c_{2} e^{\frac{t}{4}}

y^{^{''}} + 5 y^{^{'}} - 3 y = 0

a^{2} \cdot \frac{d ^{2} y}{d x ^{2}} + a \cdot \frac{d y}{d x} - c \cdot y = 0

y^{^{''''}} + 8 y^{^{''}} - 4 y^{^{'''}} - 16 y^{^{'}} + 16 y = 0

10 y^{^{''}} + 30 y^{^{'}} + 20 y = 0

\frac{d ^{2} y}{d x ^{2}} - 14 \frac{d y}{d x} + 49 y = 0