y^{''}-y^'-12y=-5e^{3x},y(0)=0,y^'(0)=0

Lösung

y^{^{''}} - y^{^{'}} - 12 y = - 5 e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{5}{42} - \frac{5}{6}) e^{4 x} - \frac{5}{42} e^{- 3 x} + \frac{5}{6} e^{3 x}

Schritte zur Lösung

y^{''} (x) - y^{'} (x) - 12 y = - 5 e^{3 x}

Solve linear ODE:

y = (\frac{5}{42} - \frac{5}{6}) e^{4 x} - \frac{5}{42} e^{- 3 x} + \frac{5}{6} e^{3 x}

y = (\frac{5}{42} - \frac{5}{6}) e^{4 x} - \frac{5}{42} e^{- 3 x} + \frac{5}{6} e^{3 x}

x^{^{''}} + 16 x^{^{'}} = 144, x (0) = 5, x^{^{'}} (0) = 12

y^{^{''}} - 90 y = sin (2 x)

y^{^{''}} + 25 \cdot y = \frac{1}{sin ( 5 \cdot x )}

y^{^{''}} + y = 3 sin (2 x) + 5 x cos (2 x)

- 6 y^{^{''}} + 5 y^{^{'}} - 1 y = 1 t^{2} - 1 t + 6 e^{2} t