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

The solution for x^{''}+2x^'+x=e^{-2t},x(0)=2 is x=e^{-t}+c_{2}e^{-t}t+e^{-2t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 13 y = cos (3 x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 x + 7

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = sin (3 x)

y^{^{''}} - 5 y^{^{'}} + 4 y = cos (x)

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