(d^2x)/(dt^2)+53/49 (dx)/(dt)+49x=0

Lösung

\frac{d ^{2} x}{d t ^{2}} + \frac{53}{49} \frac{d x}{d t} + 49 x = 0

x = e^{- \frac{53 t}{98}} (c_{1} cos (\frac{467787 t}{98}) + c_{2} sin (\frac{467787 t}{98}))

Schritte zur Lösung

\frac{d ^{2} x}{d t ^{2}} + \frac{53}{49} \frac{d x}{d t} + 49 x = 0

Solve linear ODE:

x = e^{- \frac{53 t}{98}} (c_{1} cos (\frac{467787 t}{98}) + c_{2} sin (\frac{467787 t}{98}))

x = e^{- \frac{53 t}{98}} (c_{1} cos (\frac{467787 t}{98}) + c_{2} sin (\frac{467787 t}{98}))

y^{^{''}} + 18 y^{^{'}} + 117 y = 0, y (0) = 6, y^{^{'}} (0) = 3

4 y^{^{'}} + 9 y^{^{''}} = - 5 y, y (0) = 2, y^{^{'}} (0) = 0

20 y^{^{''}} + 27 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 4

y^{^{''}} + 49 π^{2} y = 0

y^{^{''}} + 6 y^{^{'}} = 0, y (0) = - 4, y^{^{'}} (0) = 6