4y^{''}-16y^'+17y=0

Lösung

4 y^{^{''}} - 16 y^{^{'}} + 17 y = 0

y = e^{2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Schritte zur Lösung

4 y^{''} (t) - 16 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{2 t} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y (0) = 3

y^{^{''}} + (a - 1) y^{^{'}} + b y = 0

8 y^{^{''}} - 9 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = \frac{1}{8}

y^{^{''''}} - 2 y^{^{'''}} - 10 y^{^{''}} + 6 y^{^{'}} + 45 y = 0

16 \frac{d ^{4} y}{d x ^{4}} + 16 \frac{d ^{2} y}{d x ^{2}} + 4 y = 0