Question

Решить диффуры
$(xy'-y)^2=y'^2-\frac{2yy'}{x}+1$

Answer 1

$[z=\dfrac{y}{x}=>z'=\dfrac{y'x-y}{x^2}=>y'=z'x+z]\\ (z'x^2)^2=(z'x+z)^2-2z(z'x+z)+1\\ z'^2x^4=z'^2x^2-z^2+1\\ z'^2=\dfrac{1-z^2}{x^4-x^2}=>z'=\pm\sqrt{\dfrac{1-z^2}{x^4-x^2}}\\ \int \dfrac{dz}{\sqrt{1-z^2}}=\pm \int \dfrac{dx}{\sqrt{x^4-x^2}}\:\:\:(*)\\ arcsinz=\pm arctg\sqrt{x^2-1}+C_1\\ z=\pm sin(arctg\sqrt{x^2-1}+C)\\ y=\pm x\cdot sin(arctg\sqrt{x^2-1}+C)$

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$(*)\:\:\:\:\:\:\pm \int \dfrac{dx}{\sqrt{x^4-x^2}}=[\sqrt{x^2-1}=t=>dt=\dfrac{xdx}{\sqrt{x^2-1}}]=\pm\int\dfrac{dt}{t^2+1}=\pm arctgt+C=\pm arctg\sqrt{x^2-1}+C$