Solve $y' = x + y$ I am suppose to use the substitution of $u = x + y$
$y' = x + y$
$u(x) = x + y(x)$
I actually forget the trick to this and it doesn't really make much sense to me. I know that I need to get everything in a variable with x I think but I am not sure how to manipulate the problem according to mathematical rules that will make sense. Also I know that at some point I will get an integral or something and that I have no idea how to do that with multiple variables.
 A: Well, if $u = x + y$, then $y = u - x$. Take the derivative to both sides and we get 
$$ y' = u' - 1 $$
set this equal to the right hand side of our differential equation
$$ u' - 1 = x + y $$
But our substitution is $u=x+y$, so the right hand side simplifies becoming
$$ u' - 1 = u$$
thus we get a differential equation
$$ u' = 1 + u. $$
This can be solved, then we plug it back into the substitution to solve for $y$.
A: $$y'=x+y$$
Then we let $u=x+y$
This gives $u'=1+y'$, so that the equation becomes
$$u'-1=u$$
$$u'-u=1$$
Can you solve that for $u$? 
Hint $(e^x-1)'=e^x$
Moving on with the solution:
$$\frac{du}{dx}-u=1$$
$$\frac{du}{dx}=1+u$$
And the classic abuse in DE's
$$\frac{du}{u+1}=dx$$
Now 
$$\int\frac{du}{u+1}=\int dx$$
$$\log(u+1)=x+C$$
We take logarithms
$$u+1=e^{x+C}$$
We use the property of the exponential function $f(x+y)=f(x)f(y)$
$$u+1=e^C e^x$$
Here $K=e^C$
$$y+x+1=Ke^x$$
$$y=K e^x-x-1$$
A: 
I actually forget the trick to this and it doesn't really make much sense to me


Also I know that at some point I will get an integral or something and that I have no idea


This does not appear to be a seperable equation and that is all I know how to do.

We've all been there.

I am suppose to use the substitution of u=x+y

Using the given substitution is no biggie. Honestly, it's been close to a decade since I first solved this DE. And no surprises, I don't remember any bag of tricks here.
If you want the solution using the substitution $u=x+y$ by Tamaroff.
Since this is a linear ODE, you can use Laplace transform to solve this like it's a bunch of algebra.
$$y'(x)-y(x)=x$$
$$\mathcal{L}\left[y'(x)-y(x)\right]=\mathcal{L}\left[x\right]$$
Let $$\mathcal{L}\left[y(x)\right]=Y(s)$$
Then $$\mathcal{L}\left[y'(x)\right]=sY(s)-y(0)$$
Replacing them back,
$$sY-y(0)-Y=\frac{1}{s^2}$$
Assume $$y(0)=k$$
Then $$Y(s-1)=k+\frac{1}{s^2}$$
$$Y=\frac{k}{s-1}+\frac{1}{s^2(s-1)}$$
After some partial fraction decomposition,
$$Y=\frac{k}{s-1}+\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}$$
Taking the inverse Laplace,
$$\mathcal{L}^{-1}\left[Y\right]=\mathcal{L}^{-1}\left[\frac{k}{s-1}+\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}\right]$$
$$y(x)=(k+1)e^x-1-x$$
For a general solution, $k+1=K$, another constant.
Hence,
$$y(x)=Ke^x-x-1$$
