What is the solution for 2x^2y^'=y^'+6xe-y ?

The solution for 2x^2y^'=y^'+6xe-y is

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2x^2y^'=y^'+6xe-y

2 x^{2} y^{^{'}} = y^{^{'}} + 6 x e - y

y = e^{\frac{l n ( 2 x + 1 ) - l n ( 2 x - 1 )}{2 2}} \cdot \int \frac{6 e x ( ( 2 x - 1 ) ^{1 - 22} ) ^{\frac{1}{2 2}}}{( 2 x + 1 ) ^{\frac{1 + 2 2}{2 2}}} d x

Solution steps

2 x^{2} y^{'} (x) = y^{'} (x) + 6 x e - y

Solve linear ODE:

y = e^{\frac{l n ( 2 x + 1 ) - l n ( 2 x - 1 )}{2 2}} \cdot \int \frac{6 e x ( ( 2 x - 1 ) ^{1 - 22} ) ^{\frac{1}{2 2}}}{( 2 x + 1 ) ^{\frac{1 + 2 2}{2 2}}} d x

y = e^{\frac{l n ( 2 x + 1 ) - l n ( 2 x - 1 )}{2 2}} \cdot \int \frac{6 e x ( ( 2 x - 1 ) ^{1 - 22} ) ^{\frac{1}{2 2}}}{( 2 x + 1 ) ^{\frac{1 + 2 2}{2 2}}} d x

\frac{d}{d x} (- x^{5} + 3 x^{4} - 2 x^{3} - 2 x^{2} + 3 x - 1)

t^{5} e^{- t}

x \to 0 - lim (x)

f (x) = x^{2} + 2 x, g (x) = x + 2

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

What is the solution for 2x^2y^'=y^'+6xe-y ?
The solution for 2x^2y^'=y^'+6xe-y is