What is the solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=8 ?

The solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=8 is y= 160/3 e^{-(3t)/5}-160/3 e^{-(3t)/4}

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20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=8

20 y^{^{''}} + 27 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 8

y = \frac{160}{3} e^{- \frac{3 t}{5}} - \frac{160}{3} e^{- \frac{3 t}{4}}

Solution steps

20 y^{''} (t) + 27 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = \frac{160}{3} e^{- \frac{3 t}{5}} - \frac{160}{3} e^{- \frac{3 t}{4}}

y = \frac{160}{3} e^{- \frac{3 t}{5}} - \frac{160}{3} e^{- \frac{3 t}{4}}

3 y^{^{''}} + 2 y^{^{'}} + \frac{y}{4} = 0

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

y^{^{''}} + y = 0, y (0) = 19, y^{^{'}} (0) = 36

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

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0

What is the solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=8 ?
The solution for 20y^{''}+27y^'+9y=0,y(0)=0,y^'(0)=8 is y= 160/3 e^{-(3t)/5}-160/3 e^{-(3t)/4}