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

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

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

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

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

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

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = \frac{e ^{2 x}}{1 + e ^{x}}

y^{^{''}} + y^{^{'}} = (3 t^{2} + 2 t) e^{t}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 3 sin (2 x)

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

y^{^{''}} + x + y = 0

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