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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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up vote 7 down vote accepted


Then we let $u=x+y$

This gives $u'=1+y'$, so that the equation becomes



Can you solve that for $u$?

Hint $(e^x-1)'=e^x$

Moving on with the solution:



And the classic abuse in DE's



$$\int\frac{du}{u+1}=\int dx$$


We take logarithms


We use the property of the exponential function $f(x+y)=f(x)f(y)$

$$u+1=e^C e^x$$

Here $K=e^C$


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

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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 $y$.

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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

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