What is the solution for 20x^{''}+9x^'+x=0,x(0)=0,x^'(0)=4 ?

The solution for 20x^{''}+9x^'+x=0,x(0)=0,x^'(0)=4 is x=80e^{-t/5}-80e^{-t/4}

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20x^{''}+9x^'+x=0,x(0)=0,x^'(0)=4

20 x^{^{''}} + 9 x^{^{'}} + x = 0, x (0) = 0, x^{^{'}} (0) = 4

x = 80 e^{- \frac{t}{5}} - 80 e^{- \frac{t}{4}}

Solution steps

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

Solve linear ODE:

x = 80 e^{- \frac{t}{5}} - 80 e^{- \frac{t}{4}}

x = 80 e^{- \frac{t}{5}} - 80 e^{- \frac{t}{4}}

y^{^{'''}} + y^{^{''}} - 7 y^{^{'}} + 5 y = 0

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

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

(y^{^{''}} - a y^{^{'}}) = b y

y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = - 3, y^{^{'}} (0) = 0

What is the solution for 20x^{''}+9x^'+x=0,x(0)=0,x^'(0)=4 ?
The solution for 20x^{''}+9x^'+x=0,x(0)=0,x^'(0)=4 is x=80e^{-t/5}-80e^{-t/4}