What is the solution for 5x^2y^'=y^'+7xe-y ?

The solution for 5x^2y^'=y^'+7xe-y is

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5x^2y^'=y^'+7xe-y

5 x^{2} y^{^{'}} = y^{^{'}} + 7 x e - y

y = e^{\frac{l n ( 5 x + 1 ) - l n ( 5 x - 1 )}{2 5}} \cdot \int \frac{7 e x ( ( 5 x - 1 ) ^{1 - 25} ) ^{\frac{1}{2 5}}}{( 5 x + 1 ) ^{\frac{1 + 2 5}{2 5}}} d x

Solution steps

5 x^{2} y^{'} (x) = y^{'} (x) + 7 x e - y

Solve linear ODE:

y = e^{\frac{l n ( 5 x + 1 ) - l n ( 5 x - 1 )}{2 5}} \cdot \int \frac{7 e x ( ( 5 x - 1 ) ^{1 - 25} ) ^{\frac{1}{2 5}}}{( 5 x + 1 ) ^{\frac{1 + 2 5}{2 5}}} d x

y = e^{\frac{l n ( 5 x + 1 ) - l n ( 5 x - 1 )}{2 5}} \cdot \int \frac{7 e x ( ( 5 x - 1 ) ^{1 - 25} ) ^{\frac{1}{2 5}}}{( 5 x + 1 ) ^{\frac{1 + 2 5}{2 5}}} d x

\int x^{2} (1 - x)^{\frac{3}{2}} d x

t y^{^{'}} = 2 y - 3 t^{2} cos (4 t)

\frac{\partial}{\partial x} (\frac{1}{2} e^{2 x})

f (x) = x^{[2]} + 1, at x = 3

\int \frac{3}{cos ^{2} ( 3 x )} d x

What is the solution for 5x^2y^'=y^'+7xe-y ?
The solution for 5x^2y^'=y^'+7xe-y is