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

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

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x^{''}+4x^'+3x=1

x^{^{''}} + 4 x^{^{'}} + 3 x = 1

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

Solution steps

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

Solve linear ODE:

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

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

(D - 1)^{2} y = (e^{2 x} + sin (2 x))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} + 36 y = 10 cos (6 x) + 15 sin (6 x)

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 4 x^{2} - 3 x + 2

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

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