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

Lösung

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

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{223 t}{2}) + c_{2} sin (\frac{223 t}{2}))

Schritte zur Lösung

y^{''} (t) + y^{'} (t) + 56 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{223 t}{2}) + c_{2} sin (\frac{223 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{223 t}{2}) + c_{2} sin (\frac{223 t}{2}))

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = \frac{53}{5}

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = \frac{3}{5}, y^{^{'}} (0) = - \frac{1}{5}

\frac{d ^{4} y}{d x ^{4}} + 8 \frac{d ^{2} y}{d x ^{2}} + 6 y = 0

\frac{1}{2} x^{^{''}} + 3 x^{^{'}} + 4 x = 0, x (0) = 3, x^{^{'}} (0) = 0

y^{^{'''}} - 7 y^{^{''}} + 10 y = 0, y (0) = 7