What is the solution for x^{''}+3x^'+2x=1,x(0)=0,x^'(0)=0 ?

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

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x^{''}+3x^'+2x=1,x(0)=0,x^'(0)=0

x^{^{''}} + 3 x^{^{'}} + 2 x = 1, x (0) = 0, x^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 9 y = 54 t

y^{^{''}} + 10 y^{^{'}} + 25 y = \frac{e ^{- 5 x}}{x ^{6}}

y^{^{''}} - 4 y^{^{'}} + 8 y = 12 x e^{2 x} + 6 e^{2 x}

y^{^{''}} + y = 8 x cos (x)

(D^{2} - 6 D + 9) y = x^{2} e^{2 x}

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