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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$\begingroup$$$y'=x+y$$
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 $K=e^C$
$$y+x+1=Ke^x$$
$$y=K e^x-x-1$$
$\endgroup$ 19 $\begingroup$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 $y$.
$\endgroup$ 2 $\begingroup$I actually forget the trick to this and it doesn't really make much sense to me
Also I know that at some point I will get an integral or something and that I have no idea
This does not appear to be a seperable equation and that is all I know how to do.
We've all been there.
I am suppose to use the substitution of u=x+y
Using the given substitution is no biggie. Honestly, it's been close to a decade since I first solved this DE. And no surprises, I don't remember any bag of tricks here.
If you want the solution using the substitution $u=x+y$ by Tamaroff.
Since this is a linear ODE, you can use Laplace transform to solve this like it's a bunch of algebra.
$$y'(x)-y(x)=x$$$$\mathcal{L}\left[y'(x)-y(x)\right]=\mathcal{L}\left[x\right]$$Let $$\mathcal{L}\left[y(x)\right]=Y(s)$$Then $$\mathcal{L}\left[y'(x)\right]=sY(s)-y(0)$$
Replacing them back,$$sY-y(0)-Y=\frac{1}{s^2}$$Assume $$y(0)=k$$Then $$Y(s-1)=k+\frac{1}{s^2}$$$$Y=\frac{k}{s-1}+\frac{1}{s^2(s-1)}$$
After some partial fraction decomposition,$$Y=\frac{k}{s-1}+\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}$$Taking the inverse Laplace,$$\mathcal{L}^{-1}\left[Y\right]=\mathcal{L}^{-1}\left[\frac{k}{s-1}+\frac{1}{s-1}-\frac{1}{s}-\frac{1}{s^2}\right]$$$$y(x)=(k+1)e^x-1-x$$For a general solution, $k+1=K$, another constant. Hence,$$y(x)=Ke^x-x-1$$
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