y^{''}+2y^'+5y=5\H(t)

Lösung

y^{^{''}} + 2 y^{^{'}} + 5 y = 5 H (t)

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} cos (2 t) \cdot \int - \frac{5 e ^{t} sin ( 2 t ) H ( t )}{2} d t + e^{- t} sin (2 t) \cdot \int \frac{5 e ^{t} cos ( 2 t ) H ( t )}{2} d t

Schritte zur Lösung

y^{''} (t) + 2 y^{'} (t) + 5 y = 5 H (t)

Solve linear ODE:

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} cos (2 t) \cdot \int - \frac{5 e ^{t} sin ( 2 t ) H ( t )}{2} d t + e^{- t} sin (2 t) \cdot \int \frac{5 e ^{t} cos ( 2 t ) H ( t )}{2} d t

y = e^{- t} (c_{1} cos (2 t) + c_{2} sin (2 t)) + e^{- t} cos (2 t) \cdot \int - \frac{5 e ^{t} sin ( 2 t ) H ( t )}{2} d t + e^{- t} sin (2 t) \cdot \int \frac{5 e ^{t} cos ( 2 t ) H ( t )}{2} d t

y^{^{''}} + 3 \cdot y^{^{'}} + 2 \cdot y = x

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

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 1, x (0) = 0

y^{^{''}} - y^{^{'}} + y = 14 sin (t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = (x + 1) e^{x}