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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{4} (2 x + 4) d x

(2 x^{2} + 1) 3 (x^{2} - 4) 2

\int \frac{cos ^{2} ( x ) sin ( x )}{2} d x

x \to π lim (ln (cos (x) - 4))

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 t)

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