What is the solution for 5y^{''}+21y^'+4y=0,y(0)=1,y^'(0)=-19.2y ?

The solution for 5y^{''}+21y^'+4y=0,y(0)=1,y^'(0)=-19.2y is y=c_{1}e^{-t/5}+c_{2}e^{-4t}

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5y^{''}+21y^'+4y=0,y(0)=1,y^'(0)=-19.2y

5 y^{^{''}} + 21 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 19.2 y

y = c_{1} e^{- \frac{t}{5}} + c_{2} e^{- 4 t}

Solution steps

5 y^{''} (t) + 21 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{t}{5}} + c_{2} e^{- 4 t}

y = c_{1} e^{- \frac{t}{5}} + c_{2} e^{- 4 t}

4 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 9, y^{^{'}} (0) = 0.25

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (\frac{1}{3}) = 9, y (0) = 3

5 y^{^{''}} + 80 y = 0

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 11, y^{^{'}} (0) = - 49

4000 y^{^{''}} - 9 y^{^{'}} = 0

What is the solution for 5y^{''}+21y^'+4y=0,y(0)=1,y^'(0)=-19.2y ?
The solution for 5y^{''}+21y^'+4y=0,y(0)=1,y^'(0)=-19.2y is y=c_{1}e^{-t/5}+c_{2}e^{-4t}