y^{''}+4/3 y^'+αy=0,y(0)=1,y^'(0)=-16/3

Lösung

y^{^{''}} + \frac{4}{3} y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{16}{3}

y = e^{\frac{t ( - 2 + - 9 α + 4 )}{3}} (1 + \frac{- - 9 α + 4 - 14}{2 4 - 9 α}) - \frac{e ^{- \frac{t ( - 9 α + 4 + 2 )}{3}} ( - - 9 α + 4 - 14 )}{2 4 - 9 α}

Schritte zur Lösung

y^{''} (t) + \frac{4}{3} y^{'} (t) + α y = 0

Solve linear ODE:

y = e^{\frac{t ( - 2 + - 9 α + 4 )}{3}} (1 + \frac{- - 9 α + 4 - 14}{2 4 - 9 α}) - \frac{e ^{- \frac{t ( - 9 α + 4 + 2 )}{3}} ( - - 9 α + 4 - 14 )}{2 4 - 9 α}

y = e^{\frac{t ( - 2 + - 9 α + 4 )}{3}} (1 + \frac{- - 9 α + 4 - 14}{2 4 - 9 α}) - \frac{e ^{- \frac{t ( - 9 α + 4 + 2 )}{3}} ( - - 9 α + 4 - 14 )}{2 4 - 9 α}

l in e a r y^{^{''}} + 9 y = 0

y^{^{''}} - y = 0, y (- 1) = 5, y^{^{'}} (- 1) = - 5

9 y^{^{''}} - 12 y + 4 y = 0

2 y^{^{''}} + 36 y^{^{'}} + 163 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (1) = 1