y^{''}-10y^'+25y=e^{5t}ln(t)

Lösung

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 t} ln (t)

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{2 e ^{5 t} t ^{2} ln ( t ) - 3 e ^{5 t} t ^{2}}{4}

Schritte zur Lösung

y^{''} (t) - 10 y^{'} (t) + 25 y = e^{5 t} ln (t)

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{2 e ^{5 t} t ^{2} ln ( t ) - 3 e ^{5 t} t ^{2}}{4}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} + \frac{2 e ^{5 t} t ^{2} ln ( t ) - 3 e ^{5 t} t ^{2}}{4}

(D^{2} + 6 D + 9) y = \frac{16 e ^{- 3 x}}{x ^{2} + 1}

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

2 y^{^{''}} - 3 y^{^{'}} = 9 x^{2} + 10

y^{^{''}} - y^{^{'}} - 2 y = sin (t)

y^{^{''}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 2