8y^{''}+4y^'+5y=1,y(0)=1

Lösung

8 y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 1

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

Schritte zur Lösung

8 y^{''} (t) + 4 y^{'} (t) + 5 y = 1

Solve linear ODE:

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

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

y^{^{'''}} - 4 y^{^{'}} = cot (2 x)

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{3 x} + x e^{- 3 x}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = e^{3 x}

y^{^{''}} + 3 y^{^{'}} - 10 y = e^{x} cos (3 x)

y^{^{'''}} + y^{^{''}} - y^{^{'}} - y = 2 e^{- x}