What is the solution for ye^{-x}y^'=x ?

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

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ye^{-x}y^'=x

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} = \frac{sin ( x )}{sin ( y )}

\frac{d y}{d x} = \frac{x}{y ^{2}} + x y, y (1) = 1

y^{^{'}} - e^{- y} sec^{2} (x) = 0

\frac{d y}{d x} = (1 + 2 y - 1)^{2}

\frac{d}{d t} ((x) (t)) = 2 x

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