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

Lösung

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

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

Schritte zur Lösung

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

Solve linear ODE:

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

y^{^{''}} + \frac{4}{3} y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

y^{^{''}} + 15 y^{^{'}} = 0, y (0) = - 27, y^{^{'}} (0) = 15

625 y^{^{''}} + 729 y = 0, y (0) = 7, y^{^{'}} (0) = 3

l in e a r x^{^{''}} + 2 x^{^{'}} + 10 x = 0

f^{^{''}} (t) - 4 f^{^{'}} (t) + 3 f (t) = 0, f (0) = 3, f^{^{'}} (0) = 5