0
$\begingroup$

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.

2 Answers 2

7

$$y'=x+y$$

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 $K=e^C$

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

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

  • 0
    This does not appear to be a seperable equation and that is all I know how to do.2012-06-18
  • 1
    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.2012-06-18
  • 1
    @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.2012-06-18
  • 1
    @PeterTamaroff I know I don't know acronyms :$ what does IF stand for?2012-06-18
  • 0
    @AlexNelson "use the method of **i**ntegrating **f**actor"2012-06-18
  • 0
    I never would have gotten this on my own, was this something I was just suppose to be able to see or does it takes hours of trial and error?2012-06-18
  • 0
    Why is it $Ke^x$?2012-06-18
  • 0
    @Jordan $K$ is $e^C$ which comes from the constant of integration. I just changed it's name.2012-06-18
  • 0
    @PeterTamaroff I didn't get that when I did my problem. I don't see how you get it either, it just sort of appears.2012-06-18
  • 0
    @Jordan Well, when integrating, one makes $C$ appear, right?2012-06-18
  • 0
    @PeterTamaroff But a +C not a multiple.2012-06-18
  • 0
    Another idea is to write $v=u+1$, and $v'=v$...2012-06-18
  • 0
    @Jordan: The suggested substitution seems a bit far-fetched to me too. There are two standard methods that you could have use on this problem directly (without the substitution): integrating factor (since it's a linear first-order ODE), or homogeneous + particular solution (since it's a linear ODE with constant coefficients).2012-06-18
  • 0
    @HansLundmark The suggested substitution is quite common, since it "reduces" the equation, but it is better to know more theory on ODEs as you say to solve this with the general theorems.2012-06-22
  • 0
    @PeterTamaroff I have done this problem again and I still do not understand where the multiple of K comes from. I get $e^x + K -x -1$2012-06-22
  • 0
    @Jordan Please, read the answer in detail. I have added what $K$ stands for. The solution you propose is not the correct one. Try and verify what I obtain and what you obtain to see which one works. And remember, **read** thoroughfully, don't skim through the answer. Understand each step that is made, and you'll understand why the solution is what it is.2012-06-22
  • 0
    @PeterTamaroff I see that you raised everything by e but couldn't you have also done $e^x + e^c$ which would be the same?2012-06-22
  • 0
    @Jordan No. $e^{x+y}=e^x \times e^y$ not $e^{x+y}=e^y + e^x$. Go through your notes on the exponential and log functions.2012-06-22
  • 0
    @PeterTamaroff I am talking about before that though you have an x + c. If I wanted to multiply by 2 I can multiply each seperately like 2*x and 2*c so when I raise everything by e why can't I do $e^x + e^c$2012-06-22
6

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 $y$.

  • 0
    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.2012-06-18
  • 2
    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...2012-06-18