Question

(x³+y³)xy=x³-y³ dy/DX​

Answer 1

Answer:

You first observe that

x3+y3x2y+xy2=x2−xy+y2xy

with the condition that y≠−x . This is easier to manage, isn’t it? Now set y=xz , so that

y′=z+xz′

and the equation becomes

xz′=1−z+z2z−z=1−zz

and can be transformed into

z1−zz′=1x

Integrating both sides we obtain

−z−log|1−z|=log|x|+k

so

log|ez(1−z)|=log|1/x|+k

and finally

ez(1−z)=cx

Back substituting,

ey/xx−yx=cx

or, as well,

ey/x(x−y)=c