y^{''}+3y^'+ay=0,y(0)=1,y^'(0)=-8

解

y^{^{''}} + 3 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 8

y = e^{\frac{t ( - 3 + 9 - 4 a )}{2}} (1 + \frac{- - 4 a + 9 - 13}{2 9 - 4 a}) - \frac{e ^{\frac{t ( - 3 - 9 - 4 a )}{2}} ( - - 4 a + 9 - 13 )}{2 9 - 4 a}

解答ステップ

y^{''} (t) + 3 y^{'} (t) + a y = 0

線形 ODE を解く:

y = e^{\frac{t ( - 3 + 9 - 4 a )}{2}} (1 + \frac{- - 4 a + 9 - 13}{2 9 - 4 a}) - \frac{e ^{\frac{t ( - 3 - 9 - 4 a )}{2}} ( - - 4 a + 9 - 13 )}{2 9 - 4 a}

y = e^{\frac{t ( - 3 + 9 - 4 a )}{2}} (1 + \frac{- - 4 a + 9 - 13}{2 9 - 4 a}) - \frac{e ^{\frac{t ( - 3 - 9 - 4 a )}{2}} ( - - 4 a + 9 - 13 )}{2 9 - 4 a}