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
    @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 $y$.

  • 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