y^{''}-2y^'+2y=tcos(2t)

Lösung

y^{^{''}} - 2 y^{^{'}} + 2 y = t cos (2 t)

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{t sin ( 2 t )}{5} - \frac{t cos ( 2 t )}{10} - \frac{1}{25} sin (2 t) - \frac{11}{50} cos (2 t)

Schritte zur Lösung

y^{''} (t) - 2 y^{'} (t) + 2 y = t cos (2 t)

Solve linear ODE:

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{t sin ( 2 t )}{5} - \frac{t cos ( 2 t )}{10} - \frac{1}{25} sin (2 t) - \frac{11}{50} cos (2 t)

y = e^{t} (c_{1} cos (t) + c_{2} sin (t)) - \frac{t sin ( 2 t )}{5} - \frac{t cos ( 2 t )}{10} - \frac{1}{25} sin (2 t) - \frac{11}{50} cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{- 1}{2}} \cdot e^{x}

y^{^{''}} + 2 y^{^{'}} + y = e^{-} x ln (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 7 x

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

(D - 1) (D - 2) (D - 3) y = e^{x}