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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial z} (x y^{2} e^{y})

\frac{d}{d x} y = (3 - e^{- x})^{2}

\int \frac{1}{4 x - 10} d x

\frac{\partial}{\partial x} (sin^{2} (θ))

x = - 5, x = 1, y = 2 x^{2} + 4, y = 0

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