y^{'''}-6y^{''}+12y^'-8y=te^3+1

Lösung

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = t e^{3} + 1

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + c_{3} t^{2} e^{2 t} - \frac{e ^{3} t}{8} - \frac{2 + 3 e ^{3}}{16}

Schritte zur Lösung

y^{'''} (t) - 6 y^{''} (t) + 12 y^{'} (t) - 8 y = t e^{3} + 1

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + c_{3} t^{2} e^{2 t} - \frac{e ^{3} t}{8} - \frac{2 + 3 e ^{3}}{16}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + c_{3} t^{2} e^{2 t} - \frac{e ^{3} t}{8} - \frac{2 + 3 e ^{3}}{16}

y^{^{''}} + y^{^{'}} - 2 y = 2 (3 x + 1) e^{x} + 20 cos (2 x)

y^{^{''}} + 12 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{t}{e ^{t}}

y^{^{''}} + y = 2 sin (2 x)

y^{^{'''}} + 49 y^{^{'}} = 5 x + 6