y^{''}-y^'-6y=e^t,y(0)=1,y^'(0)=1

Lösung

y^{^{''}} - y^{^{'}} - 6 y = e^{t}, y (0) = 1, y^{^{'}} (0) = 1

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

Schritte zur Lösung

y^{''} (t) - y^{'} (t) - 6 y = e^{t}

Solve linear ODE:

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

y = (- \frac{7}{15} + \frac{7}{6}) e^{3 t} + \frac{7}{15} e^{- 2 t} - \frac{1}{6} e^{t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2 x} (cos (x) - 4 sin (x))

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 3 + e^{\frac{x}{2}}

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 x)

y^{^{''}} - 8 y^{^{'}} + 6 y = 3 x^{2} - 8