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

The solution for x^2y^{''}-2xy^'+2y=4x^2 is y=c_{1}x^2+c_{2}x+4x^2ln(x)-4x^2

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x^2y^{''}-2xy^'+2y=4x^2

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 4 x^{2}

y = c_{1} x^{2} + c_{2} x + 4 x^{2} ln (x) - 4 x^{2}

Solution steps

x^{2} y^{''} (x) - 2 x y^{'} (x) + 2 y = 4 x^{2}

Solve linear ODE:

y = c_{1} x^{2} + c_{2} x + 4 x^{2} ln (x) - 4 x^{2}

y = c_{1} x^{2} + c_{2} x + 4 x^{2} ln (x) - 4 x^{2}

\frac{2 s + 3}{s ^{2} + 2 s + 5}

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\frac{d}{d x} (ln x^{2} + 1 + x)

\frac{\partial}{\partial x} (x^{\frac{1}{y}})

\int \frac{1}{5 x lo g _{7} ( x )} d x

What is the solution for x^2y^{''}-2xy^'+2y=4x^2 ?
The solution for x^2y^{''}-2xy^'+2y=4x^2 is y=c_{1}x^2+c_{2}x+4x^2ln(x)-4x^2