Answer to Question #163208 in Differential Equations for Leesha

Question

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

Answer 1

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

Solution:

$y=ux$ , $dy=udx+xdu$ .

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

$4x^2(1+u)dx=ux^3du$

$4\displaystyle\frac{dx}{x}=\frac{u}{1+u}du$

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

$\ln{C}+4\ln{|x|}=u-\ln{|1+u|}$

$Cx^4=\displaystyle\frac{e^u}{1+u}$

$Cx^4=\displaystyle\frac{e^\frac yx}{1+\frac yx}$

$Cx^5=\displaystyle\frac{e^\frac yx}{x+y}$

Answer: $Cx^5=\displaystyle\frac{e^\frac yx}{x+y}$ .