y^{''}+20y^'+200y=2

Lösung

y^{^{''}} + 20 y^{^{'}} + 200 y = 2

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

Schritte zur Lösung

y^{''} (t) + 20 y^{'} (t) + 200 y = 2

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

y^{^{'''}} - 3 y^{^{''}} = e^{2 x}

y^{^{''}} + y = e^{x} + x^{3}, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 4 y = x^{- 1} e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} 4 + 8 y = 8 x^{2} + 8 x + 1

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = x^{2} + x + 1