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Question Number 101350 by bobhans last updated on 02/Jul/20

Answered by bemath last updated on 02/Jul/20

(x^2 +1)(1−y^2 )dx−xy dy=0 (x^2 +1)(1−y^2 ) dx = xy dy (((x^2 +1))/x) dx = ((y dy)/(1−y^2 )) ∫ (x+(1/x)) dx = −(1/2)∫ ((d(1−y^2 ))/(1−y^2 )) (1/2)x^2 + ln (x) +c = −(1/2)ln(1−y^2 ) ⇔x^2 + 2ln (x) + ln (1−y^2 ) = C x^2 + ln(x^2 (1−y^2 )) = C

(x2+1)(1−y2)dx−xydy=0(x2+1)(1−y2)dx=xydy(x2+1)xdx=ydy1−y2∫(x+1x)dx=−12∫d(1−y2)1−y212x2+ln(x)+c=−12ln(1−y2)⇔x2+2ln(x)+ln(1−y2)=Cx2+ln(x2(1−y2))=C

Commented by bobhans last updated on 02/Jul/20

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