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})}}

\int cos^{2} (x s i) n^{4} x d x

\int 60 x^{9} sin (3 x^{10}) d x

3 x + 1, at x = 5

\frac{\partial}{\partial x} (e^{x} - 1)

\frac{d}{d x} (\frac{1}{5 x ^{\frac{4}{5}}})

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)}}