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

The solution for x^{''}+2x^'+x=e^{-2t},x(0)=2,x^'(0)=1 is x=e^{-t}+4e^{-t}t+e^{-2t}

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

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

x = e^{- t} + 4 e^{- t} t + e^{- 2 t}

Solution steps

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

Solve linear ODE:

x = e^{- t} + 4 e^{- t} t + e^{- 2 t}

x = e^{- t} + 4 e^{- t} t + e^{- 2 t}

(D - 2)^{2} y = 20 - 3 x e^{2 x}

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

y^{^{''}} + 16 y = sec (4 t)

y^{^{''}} + 4 y = 3 sin (2 x) + x^{2} - 1

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

What is the solution for x^{''}+2x^'+x=e^{-2t},x(0)=2,x^'(0)=1 ?
The solution for x^{''}+2x^'+x=e^{-2t},x(0)=2,x^'(0)=1 is x=e^{-t}+4e^{-t}t+e^{-2t}