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Consider the IVP $ye^{-x}dy+xdx=0, y(0)=1$.

The solution is given by:

$y^2-2xe^x+2e^x=-1$

$y^2=-2xe^x+2e^x-1$

$\frac{1}{2}y^2=xe^x-e^x+\frac{3}{2}$

$y^2-2xe^x+2e^x=3$