What is the solution for y^{''}+16/3 y^'+ay=0,y(0)=1,y^'(0)=-4/3 ?

The solution for y^{''}+16/3 y^'+ay=0,y(0)=1,y^'(0)=-4/3 is

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

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

y = e^{\frac{t ( - 8 + - 9 a + 64 )}{3}} (1 + \frac{- - 9 a + 64 + 4}{2 64 - 9 a}) - \frac{e ^{- \frac{t ( - 9 a + 64 + 8 )}{3}} ( - - 9 a + 64 + 4 )}{2 64 - 9 a}

Solution steps

y^{''} (t) + \frac{16}{3} y^{'} (t) + a y (t) = 0

Solve linear ODE:

y = e^{\frac{t ( - 8 + - 9 a + 64 )}{3}} (1 + \frac{- - 9 a + 64 + 4}{2 64 - 9 a}) - \frac{e ^{- \frac{t ( - 9 a + 64 + 8 )}{3}} ( - - 9 a + 64 + 4 )}{2 64 - 9 a}

y = e^{\frac{t ( - 8 + - 9 a + 64 )}{3}} (1 + \frac{- - 9 a + 64 + 4}{2 64 - 9 a}) - \frac{e ^{- \frac{t ( - 9 a + 64 + 8 )}{3}} ( - - 9 a + 64 + 4 )}{2 64 - 9 a}

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

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

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

9 \frac{d ^{2} y}{d x ^{2}} - 49 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+16/3 y^'+ay=0,y(0)=1,y^'(0)=-4/3 ?
The solution for y^{''}+16/3 y^'+ay=0,y(0)=1,y^'(0)=-4/3 is