Solve y′=ex+yy' = e^{x+y}y′=ex+y with y(0)=0y(0) = 0y(0)=0.
y=−ln(1−ex)y = -\ln(1 - e^x)y=−ln(1−ex)
y=ln(1+ex)y = \ln(1 + e^x)y=ln(1+ex)
y=−ln(2−ex)y = -\ln(2 - e^x)y=−ln(2−ex)
y=ln(2−ex)y = \ln(2 - e^x)y=ln(2−ex)