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# Solve  xdy- ydx= xy²dx                                                                                     please solve this one  and and allow users to question below 50 characters …..........

Nathan Jacob
11 Points
4 years ago
xdy=ydx + x2y2/2 – C

On integration we get
xy=yx + x2y2/2
Therefore x2y2/2=C so either x2yis a constant value which also means xy is constant. I hope this was what you wanted. :)
20 Points
4 years ago
I do not agree.
$x^{2}y+2Cy+2x=0$
This is solved as:-
xdy-ydx=x2d(y/x)
Therefore,
x2d(y/x)=xy2dx
or
$\frac{d(y/x)}{(\frac{y^{2}}{x^{2}})}=xdx$
z-2dz=xdx       [Where z=y/x]
On integration,
$\frac{-1}{z}=\frac{x^{2}}{2}+C$
On putting z=y/x
$\frac{-x}{y}=\frac{x^{2}}{2}+C$
Simplifying
$x^{2}y+2Cy+2x=0$