(x^2+y)dy/dx+2xy=0 find the general solution ?

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20 Answers

P_=2xy ,
by Level 12 User (101k points)
æp/æy=2x
by Level 12 User (101k points)
& Q_= x^2+y ,
by Level 12 User (101k points)
æQ/æx=2x :
by Level 12 User (101k points)
P,y-Q,x=0
by Level 12 User (101k points)
ln(I•F)=0
by Level 12 User (101k points)
→ I•F= 1
by Level 12 User (101k points)
Therefore the given is an exact equation so that,
by Level 12 User (101k points)
du_= Pdx+Qdy=0
by Level 12 User (101k points)
→ u(x,y)= c1
by Level 12 User (101k points)
æu/æy= Q
by Level 12 User (101k points)
→ u=x^2 y+ y^2/2+w(x) ,
by Level 12 User (101k points)
æu/æx=2xy+w'(x)_=2xy
by Level 12 User (101k points)
→ w'(x)=0
by Level 12 User (101k points)
And w(x)= c2
by Level 12 User (101k points)
Therefore, y^2/2+x^2 y+(c2-c1)=0
by Level 12 User (101k points)
→y^2+2x^2 y+c=0
by Level 12 User (101k points)
This is a quadratic in Y with discriminant
by Level 12 User (101k points)
D=(2x^2)^2 - 4c :
by Level 12 User (101k points)
Y = -x^2 ± ab•sq•rt (x^4-c)
by Level 12 User (101k points)

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