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

The solution for x^2y^{''}-2xy^'+2y=x^2 is y=c_{1}x^2+c_{2}x+x^2ln(x)-x^2

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

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

y = c_{1} x^{2} + c_{2} x + x^{2} ln (x) - x^{2}

Solution steps

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

Solve linear ODE:

y = c_{1} x^{2} + c_{2} x + x^{2} ln (x) - x^{2}

y = c_{1} x^{2} + c_{2} x + x^{2} ln (x) - x^{2}

f (x) = \frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

\int \frac{x}{x ^{4} - 1} d x

\frac{d}{d x} (- 2 e^{- x} cos (x))

y^{^{'}} + 2 y = 0, y (0) = 1

4 y^{^{''}} + 4 y^{^{'}} + 17 = 0

What is the solution for x^2y^{''}-2xy^'+2y=x^2 ?
The solution for x^2y^{''}-2xy^'+2y=x^2 is y=c_{1}x^2+c_{2}x+x^2ln(x)-x^2