16y^{''''}+40y^{''}+25y=0

Lösung

16 y^{^{''''}} + 40 y^{^{''}} + 25 y = 0

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))

Schritte zur Lösung

16 y^{(4)} (t) + 40 y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))

y = c_{1} cos (\frac{5 t}{2}) + c_{2} sin (\frac{5 t}{2}) + t (c_{3} cos (\frac{5 t}{2}) + c_{4} sin (\frac{5 t}{2}))

y^{^{''}} + 5 y^{^{'}} - 4 y = 0

16 y^{^{''}} + 56 y^{^{'}} + 49 y = 0

y^{^{''}} - 25 y^{^{'}} = 0

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{2}) = 2, y^{^{'}} (\frac{π}{2}) = 1

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