Решите однородные дифференциальные уравнения: 1)xsqrt(1+y^2)dx+ysqrt(1+x^2)dy=0...

$x\sqrt{1+y^2}dx+y\sqrt{1+x^2}dy=0|*\frac{1}{\sqrt{1+x^2}\sqrt{1+y^2}}\\\frac{d(1+y^2)}{2\sqrt{1+y^2}}=-\frac{(1+x^2)}{2\sqrt{1+x^2}}\\\sqrt{1+x^2}=-\sqrt{1+y^2}+C\\\left[ \sqrt{1+x^2}+\sqrt{1+y^2}=C;y=^+_-i \right] \\\\\\1+y^2=0\\y^2=-1\\y=^+_-i\\y'=0\\x\sqrt{1+y^2}+y\sqrt{1+x^2}y'=0\\x*0^+_-i\sqrt{1+x^2}*0=0\\0=0$

--------------------------------------------------------------------

$xyy'=y^2+x\sqrt{9x^2+4y^2}\\y=tx;y'=t'x+t\\x^2t(t'x+t)=t^2x^2+x\sqrt{9x^2+4t^2x^2}|:x^2\\tt'x+t^2=t^2+\sqrt{9+4t^2}\\\frac{tdtx}{dx}=\sqrt{9+4t^2}|*\frac{dx}{x\sqrt{9+4t^2}}\\\frac{d(9+4t^2)}{8\sqrt{9+4t^2}}=\frac{dx}{x}\\\frac{1}{4}\sqrt{9+4t^2}=ln|x|+C\\\left[ \sqrt{9+\frac{4y^2}{x^2}}-ln|x^4|=C;y=^+_-\frac{3}{2}ix \right] \\\\9+4t^2=0\\t^2=-\frac{9}{4}\\y^2=-\frac{9}{4}x^2\\y=^+_-\frac{3}{2}ix$
$x*(^+_-\frac{3}{2}ix)*(^+_-\frac{3}{2}i)=-\frac{9}{4}x^2+x\sqrt{9x^2-9x^2}\\0=0$