Differential Equations: Exact ODEs  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Differential Equations: Exact ODEs

Postby FrozenBlood on Sat Feb 07, 2009 7:03 am

This board could really, really use a Differential Equations forum since the field of differential equations is extremely vast (and it also happens to be what I'm taking right now), but since there isn't one, I'm going to post this thread here instead.

All right, so with that said, here goes. On Tuesday, I have to present the following to my class (presentations account for 10% of my grade):

Section 2.4, #28 (from Fundamentals of Differential Equations by Nagle, Saff and Snider)

For each of the following equations, find the most general function N(x, y) so that the equation is exact:

a. [y cos(xy) + e^x]dx + N(x, y)dy = 0
b. (ye^xy - 4yx^3 + 2)dx + N(x, y)dy = 0

Note: I will use a capital D to label partial derivatives instead of the squiggly D typically used (since I don't think I'm able to post it here). And, of course, integral(x)dx will of course mean the integral of x with respect to x. I'll try my best to be as detailed as possible regarding my thinking process.

Now, I know that in order for any ODE of the form M(x, y) + N(x, y)(dy/dx) to be exact, the partial derivative of M with respect to y must be equal to the partial derivative of N with respect to x (DM/Dy = DN/Dx).

For a, the first equation, M(x, y) = y cos(xy) + e^x, and so therefore DM/Dy = x cos(xy)

The next thing I did was set x cos(xy) equal to DN/Dx:

x cos(xy) = DN/Dx

Multiplying both sides by Dx gives:

DN = x cos(xy) Dx

I suppose the next step would be to integrate both sides to isolate N(x, y) (I've never really dealt with integrals containing partial derivatives before this). I tried integration by parts, letting u = x, Dv = cos(xy) Dx, v = (1/y)sin(xy), and Du = Dx. Before I go on, I suppose I should make one small note on how I integrated Dv to get v. I used u-substitution (but actually to avoid confusion, I used w as my dummy variable instead of u), letting w = xy, Dw = y Dx, and Dx = Dw/y. Replacing all of this into the integral gives me (1/y)integral[cos(w)] Dw (note that (1/y) was factored out since it's being treated as a constant). I found the integral of that to be (1/y)sin(w) + f(y), and replacing w gives me (1/y)sin(xy) + f(y) (note how I put + f(y) instead of the usual + C). But I suppose at this point, it's okay to get rid of the f(y) part since after the full integration process is complete, I'll end up with only one f(y) instead of two. So, v = (1/y)sin(xy).

All right, enough of all of that. If we take u, Dv, v and Du and plug everything into the appropriate places of the integration by parts formula, I should get:

N(x, y) = (x/y)sin(xy) - (1/y)integral[sin(xy)] Dx (note that again I factored (1/y) out of the integral since I'm integrating with respect to x, and thus all y terms are treated as constants)

integral[sin(xy)] Dx looks like another substitution to me. I'll let z = xy, Dz = y Dx, Dx = Dz/y, which gives me (1/y)integral[sin(z)] Dz. The integral of that looks to be -(1/y)cos(z) + f(y), leading to -(1/y)cos(xy) + f(y). Now I ended up with:

N(x, y) = (x/y)sin(xy) + (y^-2)cos(xy) + f(y)

But when I take the partial derivative of that function with respect to x, I get:

DN/Dx = x cos(xy) + sin(xy) - (1/y)sin(xy).

If it weren't for that y in the denominator of the third term, the sin(xy) terms would both easily cancel out and I'd be left with x cos(xy), which matches the partial derivative of M(x, y) with respect to y.

So, where did I screw up? I haven't tried working with the second equation, but I definitely will tomorrow (it's just shortly past 2 AM right now), and anyone can certainly help me on that one as well.
FrozenBlood
 
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Re: Differential Equations: Exact ODEs  TOPIC_SOLVED

Postby FrozenBlood on Mon Feb 09, 2009 10:58 pm

I'm surprised that no one here even bothered to attempt this.

Anyway, I went back, re-evaluated my work, and it turns out I screwed up the partial derivative of M with respect to y, and from there everything went wrong. This is why I shouldn't be doing math late at night--I end up making idiotic mistakes that I shouldn't be making considering the level I'm at in my math education. I re-calculated the partial derivative (forgot to apply the product rule the first time around) and this time I got DM/Dy = cos(xy) - xy sin(xy), which I'm confident is right this time.

Then I set DN/Dx = [cos(xy) - xy sin(xy)]Dx, which leads to N(x, y) = integral[cos(xy)] Dx - y * integral[x sin(xy)] Dx. The first integral term can be done with u-substitution, the second with integration by parts. Ultimately I ended up with:

N(x, y) = x cos(xy) + f(y)

Now when I take DN/Dx, I get cos(xy) - xy sin(xy), which lo and behold, exactly matches DM/Dy, so I can't be wrong this time.

For b:

DM/Dy = yxe^xy + e^xy - 4x^3

N(x, y) = y * integral(xe^xy) Dx + integral(e^xy) Dx - 4 * integral(x^3) Dx

After a combination of integration by parts, u-substitution and the power rule, I end up with:

N(x, y) = xe^xy - x^4 + f(y)

As a last check, I calculated DN/Dx to see if it equaled yxe^xy + e^xy - 4x^3. And it did. I was right.
FrozenBlood
 
Posts: 11
Joined: Wed Feb 04, 2009 11:31 pm


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