What is the solution for y^{''}+4y^'+4y=2x+5 ?

The solution for y^{''}+4y^'+4y=2x+5 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x/2+3/4

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x}{2} + \frac{3}{4}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 2 x + 5

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x}{2} + \frac{3}{4}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x}{2} + \frac{3}{4}

y^{^{'''}} - 2 y^{^{'}} + 4 y = e^{x} cos (x) + x^{2} + sin (2 x)

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

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

y^{^{'''}} + 225 y^{^{'}} = tan (15 x)

y^{^{''}} - 2 y^{^{'}} + y = - x - 2 + 2 cos (x)

What is the solution for y^{''}+4y^'+4y=2x+5 ?
The solution for y^{''}+4y^'+4y=2x+5 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x/2+3/4