Question

x^2-3y^2+2xyy'=0, y(2)=1.

Answer 1

$x^2-3y^2+2xyy'=0\\ -3\dfrac{y^2}{x^4}+2\dfrac{yy'}{x^3}=-\dfrac{1}{x^2}\\ \dfrac{-3}{x^4}*y^2+\dfrac{1}{x^3}*2yy'=-\dfrac{1}{x^2}\\ (\dfrac{1}{x^3}*y^2)'_x=-\dfrac{1}{x^2}\\ \dfrac{1}{x^3}*y^2=-\int\dfrac{1}{x^2}dx\\ \dfrac{1}{x^3}*y^2=\dfrac{1}{x}+C_1\\ y^2=x^2+C_1x^3\\y=\pm\sqrt{x^2+C_1x^3} \\ y(2)=1=>1=\pm\sqrt{2^2+C_1*2^3} =>C_1=-\dfrac{3}{8}, y>=0\\ y=\sqrt{x^2-\dfrac{3}{8}x^3}$