Answer to Question #302641 in Differential Equations for Khushbu sharma

Question

Solve y^1=(2x+2y+3)/(2x+2y-1)

Answer 1

$y'=\frac{2x+2y+3}{2x+2y-1}\\ y'=\frac{(2x+2y-1)+4}{2x+2y-1}\\ y'=\frac{2x+2y-1}{2x+2y-1}+\frac{4}{2x+2y-1}\\ y'=1+\frac{4}{2x+2y-1}\\ u(x)=2x+2y-1\\ y=\frac{1}{2}(u-2x+1)\\ y'=\frac{1}{2}(u'-2)\\ \frac{1}{2}(u'-2)=1+\frac{4}{u}\\ \frac{1}{2}u'=\frac{2u+4}{u}\\ \frac{udu}{u+2}=4dx\\ \int{\frac{udu}{u+2}}=\int4dx\\ \int\frac{u+2-2}{u+2}du=\int4dx\\ \int(1-\frac{2}{u+2})du=4x+C\\ u-2ln|u+2|=4x+C\\ 2x+2y-1-ln(2x+2y+1)^2=4x+C\\ -2x+2y-1-ln(2x+2y+1)^2=C$