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

The solution for y^{''}+3y^'+2y=2x is y=c_{1}e^{-x}+c_{2}e^{-2x}+x-3/2

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

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

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 2 x

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 5 x^{3} + 13 x^{2} + 4 x

y^{^{''''}} + 8 y^{^{''}} + 16 y = cos (3 x) + 5

y^{^{''}} - y^{^{'}} - 2 y = cos (2 t)

y^{^{''}} + 4 y = cos (9 x)

y^{^{''}} + 4 y = csc^{2} (2 x)

What is the solution for y^{''}+3y^'+2y=2x ?
The solution for y^{''}+3y^'+2y=2x is y=c_{1}e^{-x}+c_{2}e^{-2x}+x-3/2