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