y^{''}+4y^'+5y=tan(x)

Lösung

y^{^{''}} + 4 y^{^{'}} + 5 y = tan (x)

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

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 5 y = tan (x)

Solve linear ODE:

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

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

x^{^{''}} + x = t + t^{2}

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 5 y = 1 + t

y^{^{''}} + 4 y^{^{'}} + 20 y = 4 cos (2 x)

7 y^{^{''}} + 5 y^{^{'}} - y = 9

y^{^{''}} + 3 y^{^{'}} + 2.25 y = 9 t^{3} + 64, y (0) = 1, y^{^{'}} (0) = 31.5