What is the solution for x^{''}+5x^'+6x=2e^{-t},x(0)=1,x^'(0)=0 ?

The solution for x^{''}+5x^'+6x=2e^{-t},x(0)=1,x^'(0)=0 is x=e^{-2t}-e^{-3t}+e^{-t}

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x^{''}+5x^'+6x=2e^{-t},x(0)=1,x^'(0)=0

x^{^{''}} + 5 x^{^{'}} + 6 x = 2 e^{- t}, x (0) = 1, x^{^{'}} (0) = 0

x = e^{- 2 t} - e^{- 3 t} + e^{- t}

Solution steps

x^{''} (t) + 5 x^{'} (t) + 6 x = 2 e^{- t}

Solve linear ODE:

x = e^{- 2 t} - e^{- 3 t} + e^{- t}

x = e^{- 2 t} - e^{- 3 t} + e^{- t}

y^{^{''}} - y = 2 sin (x)

x^{^{''}} - x = 4 e^{t}

y^{^{''}} + y^{^{'}} - 2 y = x^{2} + 3

y^{^{'''}} + y^{^{'}} = e^{x}, y (0) = 0

y^{^{''}} + 4 y^{^{'}} = sin^{2} (x)

What is the solution for x^{''}+5x^'+6x=2e^{-t},x(0)=1,x^'(0)=0 ?
The solution for x^{''}+5x^'+6x=2e^{-t},x(0)=1,x^'(0)=0 is x=e^{-2t}-e^{-3t}+e^{-t}