Find the general solution of the equation y′=x+y+1y' = \sqrt{x+y+1}y′=x+y+1.
2x+y+1−2ln(1+x+y+1)=x+C2\sqrt{x+y+1} - 2\ln(1+\sqrt{x+y+1}) = x + C2x+y+1−2ln(1+x+y+1)=x+C
2x+y+1−2ln∣1+x+y+1∣=x+C2\sqrt{x+y+1} - 2\ln|1+\sqrt{x+y+1}| = x + C2x+y+1−2ln∣1+x+y+1∣=x+C
2x+y+1=x+C2\sqrt{x+y+1} = x + C2x+y+1=x+C
x+y+1=x2+C\sqrt{x+y+1} = \frac{x}{2} + Cx+y+1=2x+C