What is the solution for y^'=yxe^{x+2}-y ?

The solution for y^'=yxe^{x+2}-y is y=e^{e^2(e^xx-e^x)-x+c_{1}}

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y^'=yxe^{x+2}-y

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

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

Solution steps

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

Solve separable ODE:

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

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

x y^{^{''}} + 2 y^{^{'}} = 20 x^{2}

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

lo g_{10} (x)

\frac{\partial}{\partial y} (y + z)

\frac{d}{d x} ((sin (x))^{3 x})

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