Question #163208

What is the solution for homogeneous equation (2x+y)2dx=xydy


1
Expert's answer
2021-02-23T11:25:44-0500

What is the solution for homogeneous equation (2x+y)2dx=xydy

Solution:

y=uxy=ux , dy=udx+xdudy=udx+xdu .

(2x+ux)2dx=x2u(udx+xdu)(2x+ux)^2dx=x^2u(udx+xdu)

4x2(1+u)dx=ux3du4x^2(1+u)dx=ux^3du

4dxx=u1+udu4\displaystyle\frac{dx}{x}=\frac{u}{1+u}du

4dxx=(111+u)du4\displaystyle\frac{dx}{x}=(1-\frac{1}{1+u})du

lnC+4lnx=uln1+u\ln{C}+4\ln{|x|}=u-\ln{|1+u|}

Cx4=eu1+uCx^4=\displaystyle\frac{e^u}{1+u}

Cx4=eyx1+yxCx^4=\displaystyle\frac{e^\frac yx}{1+\frac yx}

Cx5=eyxx+yCx^5=\displaystyle\frac{e^\frac yx}{x+y}

Answer: Cx5=eyxx+yCx^5=\displaystyle\frac{e^\frac yx}{x+y} .


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