(y−xy2)dx+(x+x2y2)dy=0;when y(1)=1

Priscilla Johnston

Priscilla Johnston

Answered

2022-01-06

(yxy2)dx+(x+x2y2)dy=0;when y(1)=1

Answer & Explanation

movingsupplyw1

movingsupplyw1

Expert

2022-01-07Added 30 answers

Solution:
(4xy2)dx+(x+x2y2)dx=0
(4dx+xdy)(xy2dyx2y2)=0
d(xy)(xy)2[dxxdy]=0
d(xy)(xy)2dxx+dy=0
1xylnx+y=C
put 4(1)=1
111ln1+1=C10+1=C
c=0
Hence, 1xylnx+y=0
SlabydouluS62

SlabydouluS62

Expert

2022-01-08Added 52 answers

Consider the differential equation
(yxy2)dx+(x+x2y2)dy=0
ydxxy2dx+xdy+x2y2dy=0
(ydx+xdy)+(x2y2dyxy2dx)=0
d(xy)+xy(xydyydx)=0
d(xy)+x2y2dyxy2dx=0
Dividing by x2y2
dxyx2y2+dydxx=0
dxy(xy)2+dydxx=0
Integrating the equation above we get
[dxy(xy)2]+dydxx=C…..(1)
Where C is constant of integration
Now say dzz2=1z
xndx=xn+1n+1+c for n1
dxx=lnx+c
Using these identities if z=xy then the above eqn.(1) becomes
1xy+ylnx=C
Multiply throughout by xy in the above equation we get
1+xy2xylnx=Cxy
Therefore the solution of the equation is
xy2xylnx1=Cxy.
Where C is any arbitrary constant.

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