Initial value problem, first order linear ode

[Jesssa]

Hey,

We haven't properly covered this in class yet, but I am trying to study ahead using online course notes, I manage to finish a few questions but I have gotten stuck here,

The question starts by asking for the solution to the ODE:

y' = 1 - 2xy,

When I solve this using the integrating factor technique in the final line I end up with an integral of e^{x^2}dx which I feel isn't right, wolframalpha spits out something similar but the solution involves the complex error function which I have never studied before.

Moving past that problem I started on the rest of the question,

http://img194.imageshack.us/img194/6343/asdahi.jpg

From reading them it looks like b implies c is that right?

If all approximations exist on |x|≤1/2 then they would converge to the solution right?

And the initial 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

I'm not sure on how to show b) though I keep thinking I have found which part of the theorem proof to use but I contradict myself. Could anyone steer me in the right direction?

For d) in class we were given that the difference between any approximation and the solution is

$|{{\phi }_{n}}-\phi |\,\le \frac{M}{k}\frac{{{(kh)}^{n+1}}}{(n+1)!}{{e}^{kh}}\,\,\,\,\,\,where\,\,\,\left| \frac{\partial f}{\partial y} \right|\le k$

So using this,

|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

What am I doing wrong?

 

[mighty2000]

Hello there,

Thank you for reaching out with your question. It is great to see that you are taking the initiative to study ahead using online resources. I will do my best to help you with the problem you are stuck on.

Firstly, let's take a look at the ODE y' = 1 - 2xy. This is a first-order linear differential equation, which can be solved using the integrating factor technique. The integrating factor is e^(∫(-2x)dx) = e^(-x^2). So, multiplying both sides of the equation by this factor, we get:

e^(-x^2)y' + 2xe^(-x^2)y = e^(-x^2)

Now, we can rewrite the left side as the derivative of the product e^(-x^2)y:

d/dx (e^(-x^2)y) = e^(-x^2)

Integrating both sides with respect to x, we get:

e^(-x^2)y = ∫ e^(-x^2)dx = ∫ e^(-x^2)(-2x)/(-2x)dx = -1/2∫ e^(-x^2)(-2x)dx = -1/2e^(-x^2) + C

Solving for y, we get:

y = -1/2 + Ce^(x^2)

This is the general solution to the ODE. If you have initial conditions, you can use them to find the specific solution.

Moving on to the rest of the question, b) does imply c) since if the approximations exist on |x|≤1/2, then they would also exist on |x|≤h≤1/2. And if the approximations exist on |x|≤h≤1/2, then they would converge to the solution.

For d), you have correctly used the initial theorem to find an upper bound for the difference between the approximation and the solution. The only thing you are missing is that the theorem states that this upper bound is valid for all n ≥ 1. So, you need to take that into account when calculating the upper bound. Your final answer should be ≤ 0.11/n.

I hope this helps you with your question. Let me know if you need any further clarification or assistance. Keep up the