"y'\\cdot \\frac{x}{1-x^2}=xy^\\frac{1}{2}, \\quad y(0)=1\\\\\n\\frac{dy}{y^\\frac{1}{2}}=(1-x^2)dx\\\\\n\\int\\frac{dy}{y^\\frac{1}{2}}=\\int(1-x^2)dx\\\\\n2y^\\frac{1}{2}=x-\\frac{x^3}{3}+c\\\\\ny(0)=1\\implies y=1, x=0."
Then
"2=0-0+c\\implies c=2"
Solution of differential equation is
"2y^\\frac{1}{2}=x-\\frac{x^3}{3}+2"
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