What is the solution for 2yy^'=xe^x ?

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

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2yy^'=xe^x

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

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

Solution steps

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

Solve separable ODE:

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

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

f^{^{'}} (x) = \frac{1}{1 - ( f ( x ) ) ^{2}}, f (x) (g (x)) = x

y^{^{'}} = x e^{5 x - 3 y}, y (0) = 2

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

x \frac{d y}{d x} - y = y^{2}

y^{^{'}} = k - k y^{2}

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