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

Lösung

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

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

Schritte zur Lösung

x^{''} (t) + 2 x^{'} (t) + x = e^{2 t}

Solve linear ODE:

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

x = \frac{17}{9} e^{- t} + \frac{8 e ^{- t} t}{3} + \frac{1}{9} e^{2 t}

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

16 \frac{d ^{4} y}{d x ^{4}} + 24 \frac{d ^{2} y}{d x ^{2}} + 9 y = 0

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

5 y^{^{''}} + y = 0

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0