Question #302641

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

1
Expert's answer
2022-02-27T16:51:41-0500

y=2x+2y+32x+2y1y=(2x+2y1)+42x+2y1y=2x+2y12x+2y1+42x+2y1y=1+42x+2y1u(x)=2x+2y1y=12(u2x+1)y=12(u2)12(u2)=1+4u12u=2u+4uuduu+2=4dxuduu+2=4dxu+22u+2du=4dx(12u+2)du=4x+Cu2lnu+2=4x+C2x+2y1ln(2x+2y+1)2=4x+C2x+2y1ln(2x+2y+1)2=Cy'=\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


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