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

The solution for 1/4 y^{''}+y^'+y=x^2-5x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-7x+13/2

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1/4 y^{''}+y^'+y=x^2-5x

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 5 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 7 x + \frac{13}{2}

Solution steps

\frac{1}{4} y^{''} (x) + y^{'} (x) + y = x^{2} - 5 x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 7 x + \frac{13}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 7 x + \frac{13}{2}

\int e^{x} x + e^{x} d x

f (x) = \frac{3}{2 x ^{5}}

\int \frac{1}{t ^{2} - 6 t + 13} d t

\int ln (8 x + 1) d x

\frac{( 2 s + 1 )}{s ^{2} - 6 s + 7}

What is the solution for 1/4 y^{''}+y^'+y=x^2-5x ?
The solution for 1/4 y^{''}+y^'+y=x^2-5x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-7x+13/2