Answer to Question #307501 in Differential Equations for kemmyj

Question

Find a particular solution of the differential equation: ((2y-x))/((y+2x) ) ⅆy/ⅆx=1 given that y=3 when x=2

Answer 1

$\frac{2y-x}{y+2x}\frac{dy}{dx}=1, y=3, x=2.\\ \frac{dy}{dx}=\frac{y+2x}{2y-x}\\ y=xz(x), y'=xz'+z\\ xz'+z=\frac{xz+2x}{2xz-x}\\ xz'+z=\frac{z+2}{2z-1}\\ xz'=\frac{z+2}{2z-1}-z\\ xz'=\frac{-2z^2+2z+2}{2z-1}\\ \frac{2z-1}{-2z^2+2z+2}dz=\frac{dx}{x}\\ -\frac{z-\frac{1}{2}}{z^2-z-1}dz=\frac{dx}{x}\\ -\frac{z-\frac{1}{2}}{z^2-2z\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1}dz=\frac{dx}{x}\\ -\frac{z-\frac{1}{2}}{(z-\frac{1}{2})^2 -\frac{5}{4}}dz=\frac{dx}{x}\\ -\frac{d((z-\frac{1}{2})^2 -\frac{5}{4})}{(z-\frac{1}{2})^2 -\frac{5}{4}}=2\frac{dx}{x}\\ -\int\frac{d((z-\frac{1}{2})^2 -\frac{5}{4})}{(z-\frac{1}{2})^2 -\frac{5}{4}}=2\int\frac{dx}{x}\\ -\ln|(z-\frac{1}{2})^2 -\frac{5}{4}|=2\ln|x|+\ln|C|\\ \frac{1}{(z-\frac{1}{2})^2 -\frac{5}{4}}=Cx^2\\ \frac{1}{z^2-z-1}=Cx^2\\ \frac{1}{(\frac{y}{x})^2-\frac{y}{x}-1}=Cx^2\\ \frac{1}{(\frac{3}{2})^2-\frac{3}{2}-1}=4C\\ C=-1\\ \frac{1}{(\frac{y}{x})^2-\frac{y}{x}-1}=-x^2\\$