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

The solution for y^{''}+2y^'+y=x^2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+6

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 6

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 6

y = c_{1} e^{- x} + c_{2} x e^{- x} + x^{2} - 4 x + 6

\frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 30 x + 2

y = \frac{2 x + 3}{3 x - 2}

\int 9 cos (x) + 5 d x

\frac{\partial}{\partial x} (x^{0.1} y^{0.9})

x \to 0 lim (\frac{sin ( 4 x ^{2} )}{x})

What is the solution for y^{''}+2y^'+y=x^2 ?
The solution for y^{''}+2y^'+y=x^2 is y=c_{1}e^{-x}+c_{2}xe^{-x}+x^2-4x+6