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Solve-x-2-y-1-y-2-x-1-y-0-

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Question Number 13976 by tawa tawa last updated on 25/May/17
Solve: x^2 (y + 1) + y^2 (x − 1)y′ = 0
Solve:x2(y+1)+y2(x−1)y′=0
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
y^′ =−((x^2 (y+1))/(y^2 (x−1))), y=mx y^′ =−((mx+1)/(m^2 (x−1)))=−((mx−m+m+1)/(m^2 (x−1)))= =−(1/m)+((m+1)/m^2 ).(1/(1−x))⇒y+C=−(x/m)−((m+1)/m^2 )ln(1−x)⇒ y+C=−(x^2 /y)−(((y/x)+1)/(y^2 /x^2 ))ln(1−x)⇒ −y+B=(x^2 /y)+((xy+x^2 )/y^2 )ln(1−x) (B=−C) ⇒−y^3 +By^2 =yx^2 +x(x+y)ln(1−x) . (x=1⇒y=−1)⇒1+B=−1⇒B=−2 ⇒y(y^2 +2y+x^2 )+x(x+y)ln(1−x)=0 .■
y′=−x2(y+1)y2(x−1),y=mxy′=−mx+1m2(x−1)=−mx−m+m+1m2(x−1)==−1m+m+1m2.11−x⇒y+C=−xm−m+1m2ln(1−x)⇒y+C=−x2y−yx+1y2x2ln(1−x)⇒−y+B=x2y+xy+x2y2ln(1−x)(B=−C)⇒−y3+By2=yx2+x(x+y)ln(1−x).(x=1⇒y=−1)⇒1+B=−1⇒B=−2⇒y(y2+2y+x2)+x(x+y)ln(1−x)=0.◼
Commented by tawa tawa last updated on 26/May/17
God bless you sir.
Godblessyousir.

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