What is the solution for homogeneous equation (2x+y)2dx=xydy
Solution:
y=uxy=uxy=ux , dy=udx+xdudy=udx+xdudy=udx+xdu .
(2x+ux)2dx=x2u(udx+xdu)(2x+ux)^2dx=x^2u(udx+xdu)(2x+ux)2dx=x2u(udx+xdu)
4x2(1+u)dx=ux3du4x^2(1+u)dx=ux^3du4x2(1+u)dx=ux3du
4dxx=u1+udu4\displaystyle\frac{dx}{x}=\frac{u}{1+u}du4xdx=1+uudu
4dxx=(1−11+u)du4\displaystyle\frac{dx}{x}=(1-\frac{1}{1+u})du4xdx=(1−1+u1)du
lnC+4ln∣x∣=u−ln∣1+u∣\ln{C}+4\ln{|x|}=u-\ln{|1+u|}lnC+4ln∣x∣=u−ln∣1+u∣
Cx4=eu1+uCx^4=\displaystyle\frac{e^u}{1+u}Cx4=1+ueu
Cx4=eyx1+yxCx^4=\displaystyle\frac{e^\frac yx}{1+\frac yx}Cx4=1+xyexy
Cx5=eyxx+yCx^5=\displaystyle\frac{e^\frac yx}{x+y}Cx5=x+yexy
Answer: Cx5=eyxx+yCx^5=\displaystyle\frac{e^\frac yx}{x+y}Cx5=x+yexy .
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