What is the general solution of y′=exy' = e^xy′=ex?
y=ex+Cy = e^x + Cy=ex+C
y=ex+1+Cy = e^{x+1} + Cy=ex+1+C
y=ln(x)+Cy = \ln(x) + Cy=ln(x)+C
y=xex+Cy = x e^x + Cy=xex+C