What is the solution for 1/4 y^{''}+y^'+y=x^2+2x ?

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

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1/4 y^{''}+y^'+y=x^2+2x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} + 2 x

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

Solution steps

\frac{1}{4} y^{''} (x) + y^{'} (x) + y = x^{2} + 2 x

Solve linear ODE:

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

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 6

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e x - 3

y^{^{''}} + k^{2} y = e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = - 3 cos (x) sin (x)

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