What is the solution for x^{''}+3x^'+2x=1 ?

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

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x^{''}+3x^'+2x=1

x^{^{''}} + 3 x^{^{'}} + 2 x = 1

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

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = 1

Solve linear ODE:

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

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

y^{^{'}} = \frac{1}{x} y - \frac{x ^{2}}{3} y^{4}, y (1) = 1

f (x) = (9 x + 3)^{3}

cos (sin (tan (π x)))

\frac{d ^{3} y}{d x ^{3}} + 4 \frac{d y}{d x} = 0

\frac{d y}{d x} = x + 2 y, y (0) = 4

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