Answer
$$y= Ce^{x}-(1+x)$$
Work Step by Step
Step1: Given
$$\frac{dy}{dx}= x+y, and$$$$ u=x+y$$
Step2: Substitution
$$\frac{dy}{dx}=u $$
Step3: Finding the derivative of "u"
$$\frac{du}{dx}=1+\frac{dy}{dx}$$
Step4: substitution into the equation in step 2.
$$\frac{du}{dx}-1=\frac{dy}{dx}, means,$$$$\frac{du}{dx}-1=u,or ,$$$$\frac{du}{dx}=u+1
$$
Step5; We are ready to solve!
$$\frac{du}{u+1}=dx$$
Step6; Integrating both sides with respect to corresponding variable, we get :
$$\int\frac{du}{u+1}=\int dx$$
$$ln(u+1)=x+C, where, C , is constant$$
Step7; Solving for "u" by taking exponents on both sides"e"
$$e^{ln(u+1)}=e^{x+C}$$$$e^{x+C}=e^{x}\times e^{C}$$
$$e^{C}=C (constant)$$
Step8; We have:
$$ u+1=Ce^{x}$$or$$ u=Ce^{x}-1$$
Step9; Changing the variables back
$$x+y=Ce^{x}-1$$$$means , y= Ce^{x}-(1+x)$$
