What is the solution for x^{''}+4x^'+3x=1,x^'(0)=0 ?

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

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x^{''}+4x^'+3x=1,x^'(0)=0

x^{^{''}} + 4 x^{^{'}} + 3 x = 1, x^{^{'}} (0) = 0

x = - 3 c_{2} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 3 x = 1

Solve linear ODE:

x = - 3 c_{2} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

x = - 3 c_{2} e^{- t} + c_{2} e^{- 3 t} + \frac{1}{3}

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'''}} (0) = - 2

\frac{d ^{2} y}{d x ^{2}} + y = 2 e^{3 x}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{3 t}

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

(d^{2} - 4 d + 3) y = e^{3 x} cos (x)

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