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{\partial}{\partial x} (x^{3} - 5 x y)

\frac{x d y}{d x} + (3 x + 1) y = e^{- 3 x}

\int \frac{8 x}{( 2 x ^{2} + 5 ) ^{4}} d x

x \to 0 lim (\frac{sin ( 5 x )}{tan ( 6 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