Question

Решите однородные дифференциальные уравнения:
1)xsqrt(1+y^2)dx+ysqrt(1+x^2)dy=0
2)xyy'=y^2+xsqrt(9x^2+4y^2)

Answer 1

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

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

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