y^{''}-6y^'+25y=t^2+2

Lösung

y^{^{''}} - 6 y^{^{'}} + 25 y = t^{2} + 2

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{t ^{2}}{25} + \frac{12 t}{625} + \frac{1272}{15625}

Schritte zur Lösung

y^{''} (t) - 6 y^{'} (t) + 25 y = t^{2} + 2

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{t ^{2}}{25} + \frac{12 t}{625} + \frac{1272}{15625}

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + \frac{t ^{2}}{25} + \frac{12 t}{625} + \frac{1272}{15625}

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- t} cos (2 t)

y^{^{''}} + y = 5 e^{- 2 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y^{^{'}} + 18 y = 2050 e^{(2 t)} cos (10 t)

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

\frac{d ^{2} y}{d t ^{2}} + 2 y = cos (a t)