What is the solution for x^{''}+2.25x^'+0.14x=0,x^'(0)=1,x(0)=2 ?

The solution for x^{''}+2.25x^'+0.14x=0,x^'(0)=1,x(0)=2 is

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x^{''}+2.25x^'+0.14x=0,x^'(0)=1,x(0)=2

x^{^{''}} + 2.25 x^{^{'}} + 0.14 x = 0, x^{^{'}} (0) = 1, x (0) = 2

x = - \frac{( 40 1801 - ( 65 - 1801 ) ( 45 + 1801 ) ) ( 1801 + 45 ) e ^{\frac{( - 45 + 1801 ) t}{40}}}{224 1801} - \frac{65 - 1801}{1801} e^{- \frac{( 45 + 1801 ) t}{40}}

Solution steps

x^{''} (t) + 2.25 x^{'} (t) + 0.14 x = 0

Solve linear ODE:

x = - \frac{( 40 1801 - ( 65 - 1801 ) ( 45 + 1801 ) ) ( 1801 + 45 ) e ^{\frac{( - 45 + 1801 ) t}{40}}}{224 1801} - \frac{65 - 1801}{1801} e^{- \frac{( 45 + 1801 ) t}{40}}

x = - \frac{( 40 1801 - ( 65 - 1801 ) ( 45 + 1801 ) ) ( 1801 + 45 ) e ^{\frac{( - 45 + 1801 ) t}{40}}}{224 1801} - \frac{65 - 1801}{1801} e^{- \frac{( 45 + 1801 ) t}{40}}

What is the solution for x^{''}+2.25x^'+0.14x=0,x^'(0)=1,x(0)=2 ?
The solution for x^{''}+2.25x^'+0.14x=0,x^'(0)=1,x(0)=2 is