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

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

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x^{''}+4x^'+3x=4e^{-2t},x(0)=0,x^'(0)=0

x^{^{''}} + 4 x^{^{'}} + 3 x = 4 e^{- 2 t}, x (0) = 0, x^{^{'}} (0) = 0

x = 2 e^{- t} + 2 e^{- 3 t} - 4 e^{- 2 t}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 3 x = 4 e^{- 2 t}

Solve linear ODE:

x = 2 e^{- t} + 2 e^{- 3 t} - 4 e^{- 2 t}

x = 2 e^{- t} + 2 e^{- 3 t} - 4 e^{- 2 t}

4 y^{^{''}} - 20 y^{^{'}} + 25 y = 6 - 5 x + 50 x^{2}

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

y^{^{''}} - 4 y^{^{'}} + 5 y = 5 x^{2} + 2 x

y^{^{''}} - y^{^{'}} + 4 = 3 x^{2} + 4 x

y^{^{''}} - y^{^{'}} + 4 y = 5 t^{2} + 9 t + 4

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