I have the following equation (xy^{2}+x)dx+(yx^{2}+y)dy=0 and I am told it is separabl

veksetz 2022-01-18 Answered
I have the following equation
(xy2+x)dx+(yx2+y)dy=0
and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let M=xy2+x and N=yx2+y
My=2xy and Nx=2xy
M.dxxy2+x=x2y2+x22+g(y)
Partial of (x2y2+x22+g(y))xy2+g(y)
g(y)=y
g(y)=y22
the general solution then is
C=x2y22+x22+y22
Is this solution the same I would get if I had taken the Separate Equations route?
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Expert Answer

Annie Levasseur
Answered 2022-01-19 Author has 30 answers
Basically, yes. You would write
ydyy2+1=xdxx2+1
12ln(y2+1)=12ln(x2+1)+c
Multiply both sides by 2 and exponentiate:
y2+1=e2cx2+1
This is equivalent to your solution, with C=e2c12
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Chanell Sanborn
Answered 2022-01-20 Author has 41 answers
Taking the other route you get yy(x2+1)=xy2x.
Denote z=y2, hence z=2yy. So you get z+2xx2+1z=xx2+1. This gives you z=k(x)x2+1nSKSubstituting, you have k(x)=x, which implies k(x)=x22+C. Hence z=x22(x2+1)+Cx2+1. Finally, y2=z
From here you get 2y2(x2+1)+x2=C, which is the answer you got.
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