Problem:
(y2−x2)dx+xydy=0
Solution:
(y2−x2)dx+xydy=0(y2−x2)dx=−xydydydx=x2−y2xy=xy−yx
Now let z=yx,dzdx=ddx(y)1x+yddx1x=dydx1x−yx2, which implies dydx=xdzdx+z. Putting these back, we have
dydx=xy−yxxdzdx+z=1z−zxdzdx=1z−2zzdz1−2z2=dxx−14log(1−2z2)=log(x)+C1log(1−2z2)=−4log(x)+C21−2z2=C3x−41−2(yx)2=C3x−4x2−2y2=C3x−2y2=12(x2−Cx−2)
For a quick check, we have:
y2=12(x2−Cx−2)2ydydx=x+Cx−3=x+Cx−2x=x+x2−2y2x=2x−2y2xdydx=xy−yx
(y2−x2)dx+xydy=0
Solution:
(y2−x2)dx+xydy=0(y2−x2)dx=−xydydydx=x2−y2xy=xy−yx
Now let z=yx,dzdx=ddx(y)1x+yddx1x=dydx1x−yx2, which implies dydx=xdzdx+z. Putting these back, we have
dydx=xy−yxxdzdx+z=1z−zxdzdx=1z−2zzdz1−2z2=dxx−14log(1−2z2)=log(x)+C1log(1−2z2)=−4log(x)+C21−2z2=C3x−41−2(yx)2=C3x−4x2−2y2=C3x−2y2=12(x2−Cx−2)
For a quick check, we have:
y2=12(x2−Cx−2)2ydydx=x+Cx−3=x+Cx−2x=x+x2−2y2x=2x−2y2xdydx=xy−yx
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