Solve y^1=(2x+2y+3)/(2x+2y-1)
y′=2x+2y+32x+2y−1y′=2x+2y−1+42x+2y−1y′=1+42x+2y−12x+2y−1=z(x)y=12(z−2x+1)y′=12(z′−2)12(z′−2)=1+4z12z′=2z+4zzdzz+2=4dx∫zdzz+2=∫4dxz+2−2dzz+2=∫4dx∫(1−2z+2)dz=4x+Cz−2ln∣z+2∣=4x+C2x+2y−1−2ln∣2x+2y+1∣=4x+C−2x+2y−1−2ln∣2x+2y+1∣=Cy'=\frac{2x+2y+3}{2x+2y-1}\\ y'=\frac{2x+2y-1+4}{2x+2y-1}\\ y'=1+\frac{4}{2x+2y-1}\\ 2x+2y-1=z(x)\\ y=\frac{1}{2}(z-2x+1)\\ y'=\frac{1}{2}(z'-2)\\ \frac{1}{2}(z'-2)=1+\frac{4}{z}\\ \frac{1}{2}z'=\frac{2z+4}{z}\\ \frac{zdz}{z+2}=4dx\\ \int{\frac{zdz}{z+2}}=\int4dx\\ \frac{z+2-2dz}{z+2}=\int4dx\\ \int(1-\frac{2}{z+2})dz=4x+C\\ z-2ln|z+2|=4x+C\\ 2x+2y-1-2ln|2x+2y+1|=4x+C\\ -2x+2y-1-2ln|2x+2y+1|=Cy′=2x+2y−12x+2y+3y′=2x+2y−12x+2y−1+4y′=1+2x+2y−142x+2y−1=z(x)y=21(z−2x+1)y′=21(z′−2)21(z′−2)=1+z421z′=z2z+4z+2zdz=4dx∫z+2zdz=∫4dxz+2z+2−2dz=∫4dx∫(1−z+22)dz=4x+Cz−2ln∣z+2∣=4x+C2x+2y−1−2ln∣2x+2y+1∣=4x+C−2x+2y−1−2ln∣2x+2y+1∣=C
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