What is the solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 ?

The solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 is x= 3/4 e^{-2t}cos(2t)+1/4 e^{-2t}

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x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2

x^{^{''}} + 4 x^{^{'}} + 8 x = e^{- 2 t}, x (0) = 1, x^{^{'}} (0) = - 2

x = \frac{3}{4} e^{- 2 t} cos (2 t) + \frac{1}{4} e^{- 2 t}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 8 x = e^{- 2 t}

Solve linear ODE:

x = \frac{3}{4} e^{- 2 t} cos (2 t) + \frac{1}{4} e^{- 2 t}

x = \frac{3}{4} e^{- 2 t} cos (2 t) + \frac{1}{4} e^{- 2 t}

\frac{d}{d x} (cos (2 x) - 1)

\frac{d y}{d x} - \frac{1}{x} y = 0

\int_{- 2}^{3} \frac{36}{x ^{4}} d x

\int 2 ln (2 x + 1) d x

What is the solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 ?
The solution for x^{''}+4x^'+8x=e^{-2t},x(0)=1,x^'(0)=-2 is x= 3/4 e^{-2t}cos(2t)+1/4 e^{-2t}