Solve $y' = x + y$

Solution 1:

$$y'=x+y$$

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 $K=e^C$

$$y+x+1=Ke^x$$

$$y=K e^x-x-1$$

Solution 2:

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 $y$.