y^{''}+4y^'+4y=sin(x),y(0)=1,y^'(0)=0

Lösung

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 0

y = \frac{29}{25} e^{- 2 x} + \frac{11 e ^{- 2 x} x}{5} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) + 4 y = sin (x)

Solve linear ODE:

y = \frac{29}{25} e^{- 2 x} + \frac{11 e ^{- 2 x} x}{5} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y = \frac{29}{25} e^{- 2 x} + \frac{11 e ^{- 2 x} x}{5} + \frac{3}{25} sin (x) - \frac{4}{25} cos (x)

y^{^{''}} + y - 2 e^{- t} - 1 = 0

\frac{d ^{2}}{d x ^{2}} (θ (x)) + 4 θ (x) = csc^{2} (2 x)

(D^{4} + D^{2}) y = 3 x^{2} + 4 sin (x) - 2 cos (x)

y^{^{''}} + 2 y = - 18 x e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = 2 cos (x) - 2 e^{3 x} + 10