What is the solution for y^'=xy+x-2y-2,y(0)=2 ?

The solution for y^'=xy+x-2y-2,y(0)=2 is y=3e^{(x^2)/2-2x}-1

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y^'=xy+x-2y-2,y(0)=2

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

y = 3 e^{\frac{x ^{2}}{2} - 2 x} - 1

Solution steps

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

Solve separable ODE:

y = 3 e^{\frac{x ^{2}}{2} - 2 x} - 1

y = 3 e^{\frac{x ^{2}}{2} - 2 x} - 1

4 y \frac{d y}{d x} = 6 x^{2}

y^{3} \frac{d y}{d x} = 5 x^{2} y^{4} - 20 x^{2}

y \frac{d y}{d x} = x^{2}, y (0) = 1

\frac{d}{d x} (f (x)^{- 1}) x^{2} - 8 x - 9 x = - 21

y^{^{'}} = \frac{y - 1}{x}

What is the solution for y^'=xy+x-2y-2,y(0)=2 ?
The solution for y^'=xy+x-2y-2,y(0)=2 is y=3e^{(x^2)/2-2x}-1