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

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

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

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

y = - \frac{x ^{2} + 3 x + 3}{x + 1} + \frac{c _{1}}{e ^{- 1 - x + c_{1}} + e ^{- 1 - x + c_{1}} x}

Solution steps

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

Solve linear ODE:

y = - \frac{x ^{2} + 3 x + 3}{x + 1} + \frac{c _{1}}{e ^{- 1 - x + c_{1}} + e ^{- 1 - x + c_{1}} x}

y = - \frac{x ^{2} + 3 x + 3}{x + 1} + \frac{c _{1}}{e ^{- 1 - x + c_{1}} + e ^{- 1 - x + c_{1}} x}

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

\frac{d}{d x} (5 x^{3} 5 x - 2)

\int 1 - x^{2} + \frac{1}{2} x^{4} d x

x^{3} - 2 x^{2} + 3 x - 5, 2

x \to 4 lim (2 x^{2} - 11 x)

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