y^{''}-2y^'+y=0,y(0)=8,y^'(0)=3

Lösung

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = 3

y = 8 e^{t} - 5 e^{t} t

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + y = 0

Solve linear ODE:

y = 8 e^{t} - 5 e^{t} t

y = 8 e^{t} - 5 e^{t} t

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + 5^{2} y = 0

l in e a r 12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (10) = 0

y^{^{'''}} - 15 y^{^{''}} + 72 y^{^{'}} - 108 y = 0

y^{^{'''}} + (- 7) y^{^{''}} + 8 y^{^{'}} + 16 y = 0