(d^3x)/(dt^3)+x= 1/2 t^2e^t

Lösung

\frac{d ^{3} x}{d t ^{3}} + x = \frac{1}{2} t^{2} e^{t}

x = c_{1} e^{- t} + e^{\frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) + \frac{e ^{t} t ^{2}}{4} - \frac{3 e ^{t} t}{4} + \frac{3}{8} e^{t}

Schritte zur Lösung

\frac{d ^{3} x}{d t ^{3}} + x = \frac{1}{2} t^{2} e^{t}

Solve linear ODE:

x = c_{1} e^{- t} + e^{\frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) + \frac{e ^{t} t ^{2}}{4} - \frac{3 e ^{t} t}{4} + \frac{3}{8} e^{t}

x = c_{1} e^{- t} + e^{\frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) + \frac{e ^{t} t ^{2}}{4} - \frac{3 e ^{t} t}{4} + \frac{3}{8} e^{t}

y^{^{''}} - 9 y^{^{'}} + 14 y = 3 x^{2} - 5 sin (2 x) + 7 x e^{6 x}

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

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 cos (2 x) + (3 x + 2) e^{- 2 x}

2 y^{^{''}} - 4 y^{^{'}} + 4 y = e x sec (x)

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 6 + e^{\frac{x}{2}}