What is the solution for xy^'=((1-x)y+x^2+x^3),y(3)=2 ?

The solution for xy^'=((1-x)y+x^2+x^3),y(3)=2 is y=x^2-(\W_{0}((7e^3)/3)x)/(e^{x-\W_{0)((7e^3)/3)}}

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xy^'=((1-x)y+x^2+x^3),y(3)=2

x y^{^{'}} = ((1 - x) y + x^{2} + x^{3}), y (3) = 2

y = x^{2} - \frac{W _{0} ( \frac{7 e ^{3}}{3} ) x}{e ^{x - W_{0} (\frac{7 e ^{3}}{3})}}

Solution steps

x y^{'} (x) = ((1 - x) y + x^{2} + x^{3})

Solve linear ODE:

y = x^{2} - \frac{W _{0} ( \frac{7 e ^{3}}{3} ) x}{e ^{x - W_{0} (\frac{7 e ^{3}}{3})}}

y = x^{2} - \frac{W _{0} ( \frac{7 e ^{3}}{3} ) x}{e ^{x - W_{0} (\frac{7 e ^{3}}{3})}}

x \to 3 - lim (\frac{x}{∣ x ∣})

\frac{\partial}{\partial x} (2 y^{7} + 7 y^{9} x^{6})

\int 2 x (x + 1) d x

x \to 2 lim (x^{3} + 8)

\frac{d}{d x} (7 x arcsin (x))

What is the solution for xy^'=((1-x)y+x^2+x^3),y(3)=2 ?
The solution for xy^'=((1-x)y+x^2+x^3),y(3)=2 is y=x^2-(\W_{0}((7e^3)/3)x)/(e^{x-\W_{0)((7e^3)/3)}}