Solve dydx=xex−y\frac{dy}{dx} = x e^{x-y}dxdy=xex−y with the initial condition y(0)=0y(0) = 0y(0)=0. Which expression gives y(x)y(x)y(x)?
y=ln(xex−ex+2)y = \ln(x e^x - e^x + 2)y=ln(xex−ex+2)
y=ln(xex+1)y = \ln(x e^x + 1)y=ln(xex+1)
y=ln(xex−ex+1)y = \ln(x e^x - e^x + 1)y=ln(xex−ex+1)
y=ex2/2y = e^{x^2/2}y=ex2/2