Question on Approximate Integrals?

[mathnoob]

The Questions

Let y(x) be the position of a particle at time x. Suppose that we know that
the velocity of a particle satisfies the differential equation:

y'(t) = ty

with y(0) = 1. We will try to give a reasonable method to approximate some of
the positions of the particle.

a.) Integral both sides to show that:
y(x) = integral(x to 0) ty(t) dt +1

b.)In your own words, explain how a numerical technique could be set up to
approximate solutions to the differential equation;

y'' = ty

Hint: integration by parts.

3. My attempt at a solution

a.) Since y'(t) = ty, when we integrate both sides --> y(t) = (t/2)y^2. I don't see how this can turn to a y(x)...

 

[rock.freak667]

3. My attempt at a solution

a.) Since y'(t) = ty, when we integrate both sides --> y(t) = (t/2)y^2. I don't see how this can turn to a y(x)...

Integrate both sides w.r.t. t between 0 and x. [Remember, you know what y(0) is equal to]

 

[mathnoob]

What do you mean by that... I'm still confused =S

 

[abhishek ghos]

y''=ty
(ty)'=ty
let ty =z
z'=z
z(0)=0

 

[uart]

The Questions

Let y(x) be the position of a particle at time x. Suppose that we know that
the velocity of a particle satisfies the differential equation:

y'(t) = ty

with y(0) = 1. We will try to give a reasonable method to approximate some of
the positions of the particle.

a.) Integral both sides to show that:
y(x) = integral(x to 0) ty(t) dt +1

b.)In your own words, explain how a numerical technique could be set up to
approximate solutions to the differential equation;

y'' = ty

Hint: integration by parts.

3. My attempt at a solution

a.) Since y'(t) = ty, when we integrate both sides --> y(t) = (t/2)y^2. I don't see how this can turn to a y(x)...

Just checking, are parts a) and b) meant to be two different functions or is that a typo?

a) y' = ty

b)y'' = ty

I think you're supposed to be looking a ways of numerically approximating these integrals right.

The integral of part a) is trivial to solve and has a nice closed form solution in standard maths functions, so it's not really such a good example case for numerical methods. In any case you can still use numerical methods on it, even though it's not the best approach for this particular problem.

The most basic numerical technique to solve equations of the form y' =f(x,y) is to iterate :

$$y_{k+1} = y_k + \Delta f(x_k,y_k)$$

Where $\Delta = x_{k+1}-x_k$, that is Delta is the x inter-sample interval.

You start with $y_0$ set to your initial condition and select a value of Delta which is suitably small for the desired level of approximation. You then just directly calculate y1, y2, y3, which actually correspond to $y(\Delta), y(2\Delta), y(3\Delta)$ etc.

 

[uart]

To handle higher order derivatives, for example problems of the form y'' =f(x,y), you just introduce new variables for each of the derivatives.

For example in this case let [itex]y' = v[/tex] and then iterate the following set of equations.<br /> <br /> $$y_{k+1} = y_k + \Delta v_k$$<br /> $$v_{k+1} = v_k + \Delta f(x_k,y_k)$$<br /> <br /> As before you start with $y_0$ and $v_0$ set to your initial conditions (on y and y' respectively) and calculate the future values step by step (iteratively).BTW. This is a very simple method (roughly equivalent to approximating area using Reimanm left sum). Much better methods are available, the "Runge Kutter" method for example is more complicated but much more accurate (roughly like Simpsons rule compared with Riemann left sum).[/itex]