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

The solution for 1/4 y^{''}+y^'+y=x^2-3x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-5x+9/2

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1/4 y^{''}+y^'+y=x^2-3x

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 5 x + \frac{9}{2}

Solution steps

\frac{1}{4} y^{''} (x) + y^{'} (x) + y = x^{2} - 3 x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 5 x + \frac{9}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 5 x + \frac{9}{2}

y^{'} = \frac{e ^{3 \cdot x} + 3 \cdot x \cdot y + y}{x}

\int \frac{1}{2 - 7 x} d x

sin ((π x)^{9})

[x^{2} (x^{2} + 5 x)]^{6}

What is the solution for 1/4 y^{''}+y^'+y=x^2-3x ?
The solution for 1/4 y^{''}+y^'+y=x^2-3x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-5x+9/2