i have dy/dx=x+y
y0=1
y1=1+x+(x^2)/2
y2=1+x+(x^2)/2+(x^3)/6
what is the limit of the series yn
?
Two comments:
1. It looks an awful lot like the exponential series.
2. Theoretically, assuming certain smoothness conditions are met, which it looks like they are, the series should converge to the solution of the DE. You're essentially doing Picard iterations there, if I'm not mistaken.
So, just to be clear, you're setting up the following iteration scheme:
$\displaystyle \displaystyle y_{n+1}(x)=\int(x+y_{n}(x))\,dx,$ right? If so, I don't get what you get. I'm seeing
$\displaystyle y_{0}=1,$
$\displaystyle y_{1}=\displaystyle\int(x+1)\,dx=x+\frac{x^{2}}{2} ,$
$\displaystyle \displaystyle y_{2}=\int\left(x+x+\frac{x^{2}}{2}\right)\,dx=x^{ 2}+\frac{x^{3}}{6},\dots$
Is there an initial condition or something I should know about?
Ok. So y(0) = 1 implies that
$\displaystyle y_{0}=1,$
$\displaystyle y_{1}=\displaystyle\int(x+1)\,dx=1+x+\frac{x^{2}}{ 2},$
$\displaystyle y_{2}=\displaystyle\int\left(x+1+x+\frac{x^{2}}{2} \right)\,dx=1+x+x^{2}+\frac{x^{3}}{6},$
$\displaystyle y_{3}=\displaystyle\int\left(x+1+x+x^{2}+\frac{x^{ 3}}{6}\right)dx=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6 }+\frac{x^{4}}{24},\dots$
That's what you got, I believe. The sequence $\displaystyle \{y_{n}\}$ converges to $\displaystyle e^{x}.$ To convince yourself of that fact, just expand $\displaystyle e^{x}$ in a Taylor series about 0, and you'll see that it's the exact same series as you're getting here.
How rigorous a proof does this require?
We have the first order linear DE...
$\displaystyle \displaystyle y^{'} = y + x\ \ ,\ y(0)=1$ (1)
... and we have 'forgotten' the procedure to solve it... no problem because we can use an alternative approach!...
Let's suppose that y(x) is analytic in x=0 so that is...
$\displaystyle \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)
But the $\displaystyle y^{(n)} (0)$ can be derived from (1) as follows...
$\displaystyle \displaystyle y^{(0)} (0) = 1$
$\displaystyle \displaystyle y^{(1)}(x)= y+x \implies y^{(1)}(0)= 1$
$\displaystyle \displaystyle y^{(2)}(x) = 1 + y^{(1)}(x) = 1+x+y \implies y^{(2)}(0)= 2$
$\displaystyle \displaystyle y^{(3)}(x) = 1+x+y \implies y^{(3)}(0)= 2$
$\displaystyle \dots$
$\displaystyle \displaystyle y^{(n)}(x) = 1+x+y \implies y^{(n)}(0)= 2$
$\displaystyle \dots$
... so that is...
$\displaystyle \displaystyle y(x)= 1 + x + 2\ \sum_{n=2}^{\infty} \frac{x^{n}}{n!} = 2\ e^{x} -1 -x$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$