# 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.

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$$y'=x+y$$

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 $K=e^C$

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

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

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This does not appear to be a seperable equation and that is all I know how to do. – user138246 Jun 18 '12 at 0:36
Well, you're supposed to use the method of integrating factor, so multiply both sides by $\exp(x)$, and then you get $(u'-u)\exp(x)=\exp(x)$. The left hand side is $(u\cdot\exp(x))'$, and you can integrate everything nicely. – Alex Nelson Jun 18 '12 at 0:38
@AlexNelson No one is supposed to do anything. One can choose a variety of paths. In this case, noting that $e^x-(e^x-1)=1$ is one simple solution. I don't think he knows about IFs. – Pedro Tamaroff Jun 18 '12 at 0:42
@PeterTamaroff I know I don't know acronyms :$what does IF stand for? – Alex Nelson Jun 18 '12 at 0:43 @AlexNelson "use the method of integrating factor" – Pedro Tamaroff Jun 18 '12 at 0:46 Well, if$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$y$. - I still get a funtion I do not know how to work with. I do not know how to find the integral of something without a dy or dx or whatever. – user138246 Jun 18 '12 at 0:41 Well, you're supposed to divide both sides by$1+u$, right? You get$u'/(1+u)=1$. Integrate both sides with respect to$x$, so you get$\int (1+u)^{-1}\,du = \int dx$. Performing this integral gives you$\ln(1+u) = x-x_{0}\$. Exponentiate both sides, and you're almost done... – Alex Nelson Jun 18 '12 at 0:42