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

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

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y^{''}+3y^'+2y=3

y^{^{''}} + 3 y^{^{'}} + 2 y = 3

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 2

y^{^{''}} + 2 y^{^{'}} + y = ln (x) \cdot e^{- x}

x^{^{''}} - 14 x^{^{'}} + 49 x = 5 t e^{7 t}

y^{^{''}} + 16 y = cos (4 x)

y^{^{''}} - 5 y^{^{'}} - 36 y = x - 8

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