Integral equations
By Brad Rodgers (P1930) on Monday, March 19, 2001 - 10:54 pm :

Is there a way to solve integral equations such as


x=ò1x(1+dy/dx) dx
I can't seem to think of a way to solve this, or any others like it.

Thanks,

Brad
By Michael Doré (Md285) on Monday, March 19, 2001 - 11:08 pm :

We need to clarify what the equation means. If you're integrating dx then x is a variable so it is not allowed to appear in the integration limits. Nor is it then allowed to appear on the LHS of the equation. I think what you mean is this:

Find all functions y(x) which satisfy (for every fixed x):

x=ò1x(1+y ' (t))dt

where y ' (t) represents dy/dx at x=t.

If this is what you mean, then the thing to do is differentiate each side with respect to x. (We're allowed to do this as the equation is an identity in x.) The RHS comes to 1+y ' (x) by the fundamental theorem of calculus. We get:

1=1+y ' (x)

So y=const is the only solution.

By Brad Rodgers (P1930) on Monday, March 19, 2001 - 11:31 pm :

Actually, that particular example is a bit easy, so here's another one that might prove more useful


x=ò1x(1+[dy/dx]2)1/2 dx
By Brad Rodgers (P1930) on Monday, March 19, 2001 - 11:34 pm :

Oops, hadn't yet seen your post, your interpretation is what I mean, so to rephrase the second question correctly


x=ò1x(1+[dy/dt]2)1/2dt
By Michael Doré (Md285) on Monday, March 19, 2001 - 11:56 pm :

Same idea. You can differentiate both sides, and using the fundamental theorem of calculus:

1=(1+y ' (x)2)1/2

So y ' (x)2=0 and again y=const is the only possibility.

Are you happy with the fundamental theorem of calculus step? The fact we need to use is that if:

z(x)=ò0x f(t) dt

then z ' (x)=f(x) provided f is nicely behaved.

By Brad Rodgers (P1930) on Tuesday, March 20, 2001 - 12:32 am :

Sorry, hadn't really read that last post either, now that I understand that, I think that I see how to do all of these. Since this thread is mainly devoted to differential equations, I'll go ahead and post the original question I wanted to apply this to

Solve:


dy/dx=(y+ò100x 2/5(1+[dy/dx]2)1/2 dx)/x
Thus far, I've gotten using the method above and ordinary differential solving methods,

dy/dx=tan(-4 /25x +C)

This doesn't seem to integrate easily. Are there other ways than just substituting dy/dx=z?

For that bit about the fundemental theorem of calculus, it doesn't seem intuitional that a problem involving integrating from 0 to x would have the same answer as if I were integrating from 100 to x. How can this be?

Thanks,

Brad
By Michael Doré (Md285) on Tuesday, March 20, 2001 - 12:47 am :

If you let p(x) be the integral of f(t) from 100 to x and let q(x) be the integral of f(t) from 0 to x then:

q(x)=p(x)+ò0100 f(t)dt

The last term on the RHS is a constant. So differentiating we get:

q ' (x)=p ' (x) as required.

By Michael Doré (Md285) on Tuesday, March 20, 2001 - 12:53 am :

For the problem at the start of your message, once again we need to separate the integration variable from the constant x.