What is the solution for 11x^{''}+112.5x=5e^{-t},x(0)=0,x^'(0)=0 ?

The solution for 11x^{''}+112.5x=5e^{-t},x(0)=0,x^'(0)=0 is

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11x^{''}+112.5x=5e^{-t},x(0)=0,x^'(0)=0

11 x^{^{''}} + 112.5 x = 5 e^{- t}, x (0) = 0, x^{^{'}} (0) = 0

x = - \frac{10}{247} cos (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10 1.25 E 14}{247 1.27841 E 15} sin (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10}{247} e^{- t}

Solution steps

11 x^{''} (t) + 112.5 x = 5 e^{- t}

Solve linear ODE:

x = - \frac{10}{247} cos (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10 1.25 E 14}{247 1.27841 E 15} sin (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10}{247} e^{- t}

x = - \frac{10}{247} cos (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10 1.25 E 14}{247 1.27841 E 15} sin (\frac{1.27841 E 15}{1.25 E 14} t) + \frac{10}{247} e^{- t}

What is the solution for 11x^{''}+112.5x=5e^{-t},x(0)=0,x^'(0)=0 ?
The solution for 11x^{''}+112.5x=5e^{-t},x(0)=0,x^'(0)=0 is