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Hey,
So I've been trying to solve this ode,
I got stuck at the point where I end up integrating ex^2dx
The ODE is
y'=1-2xy with y(0)=0
I double checked with wolframalpha but it spits out a solution involving the complex error function (which I have never studied before)
Is there some other way to solve this ODE?
$\displaystyle \displaystyle \begin{align*} e^{x^2} \end{align*}$, while the integral does exist, it can not be written in terms of elementary functions. So you will be able to leave your solution in terms of $\displaystyle \displaystyle \begin{align*} \int{e^{x^2}\,dx} \end{align*}$.Quote:
Originally Posted by Jesssa
Oh cool thanks!
Hey, I also have this list of other questions to practice on using afew odes including the one above,
For the one above I've been having trouble showing b,
http://img194.imageshack.us/img194/6343/asdahi.jpg
From reading them it looks like b implies c aswell is that right?
If all approximations exist on |x|≤1/2 then they would converge to the solution right?
And the first theorm in my notes states that if |x|≤h≤a then there exists a unique solution y=phi, where h = Min(a, b/M) and |f(x,y)|=|1-2xy|≤M
But I've been having trouble showing b, would you be able to point me in the right direction?
I worked out part d) for one of the other odes following an example in my notes but for this one i get, using the below relation
$\displaystyle |{{\phi }_{n}}-\phi |\,\le \frac{M}{k}\frac{{{(kh)}^{n+1}}}{(n+1)!}{{e}^{kh}} \,\,\,\,\,\,where\,\,\,\left| \frac{\partial f}{\partial y} \right|\le k$
|f(x,y)|=|1-2xy|≤1 + 2 (1/2) 1 = 2 = M
|df(x,y)/dy|=|1-2x|≤2=k
and h = 1/2
Subbing it all into the equation i get ≤0.11
Is there something I am doing incorrectly? I took this approach before and got <0.01
