What is the solution for xy^'-2y=x^2-2x-1 ?

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

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xy^'-2y=x^2-2x-1

x y^{^{'}} - 2 y = x^{2} - 2 x - 1

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

Solution steps

x y^{'} (x) - 2 y = x^{2} - 2 x - 1

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (cot (x y))

\frac{1}{( 1 - x )}, 2

\int \frac{x ^{2} - 19}{x ^{3} - 2 x ^{2} + x} d x

\int 0.2 x d x

\frac{d x}{d t} = \frac{7 x ^{2} + 3 t 8 t ^{2} + x ^{2}}{7 t x}

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