y^{''}-10y^'+25y=(1.5e^{5t})/(t^2+1)

Lösung

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{1.5 e ^{5 t}}{t ^{2} + 1}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - 0.75 e^{5 t} ln (t^{2} + 1) + 1.5 e^{5 t} t arctan (t)

Schritte zur Lösung

y^{''} (t) - 10 y^{'} (t) + 25 y = \frac{1.5 e ^{5 t}}{t ^{2} + 1}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - 0.75 e^{5 t} ln (t^{2} + 1) + 1.5 e^{5 t} t arctan (t)

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - 0.75 e^{5 t} ln (t^{2} + 1) + 1.5 e^{5 t} t arctan (t)

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 e^{x} + x

x^{^{''}} - 3 x^{^{'}} - 4 x = 8 e^{t} cos (2 t)

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = e^{6 x}

y^{^{''}} - 2 y^{^{'}} + y = t^{- 1} e^{t}