From: Doug Magnoli <dmagnoli@home.com>
Subject: Re: STUDENT NEEDS HELP  :o(
Date: Sun, 08 Apr 2001 22:42:02 GMT
Newsgroups: sci.math.num-analysis
Summary: Elementary reduction of homogeneous ODE

First of all, this problem is better posted in alt.alg.help or
alt.math.undergrad.  Second, if STUDENT NEEDS HELP :o(, then STUDENT NEEDS to
learn to post what STUDENT has done on a problem and where STUDENT is stuck.

M dx + N dy = 0

Set x = r cos q, y=r sin q

M dx = M (dx/dr) dr + M (dx/dq) dq = M cos q dr - M r sin q dq

N dy = N(dy/dr) dr + N(dy/dq) dq = N sin q dr + N r cos q dq

M dx = - N dy

M cos q dr - M r sin q dq = -N sin q dr - N r cos q dq

M cos q dr + N sin q dr = -N r cos q dq + M r sin q dq

(M cos q + N sin q) dr = r (-N cos q + M sin q) dq

dr / r = (-N cos q + M sin q) dq / (M cos q + N sin q)   (*)

----

(x+y) dx - x dy = 0

M = x+y = r(cos q + sin q)
N = -x = -r cos q

Substituting into (*), we get:

dr / r = (r cos^2 q + r  sin q (cos q + sin q)) dq / (r (cos q + sin q) cos q - r
cos q sin q)

dr/r = (cos^2 q + sin q cos q + sin^2 q) dq / (cos^2 q + sin q cos q - cos q sin
q)

 = (1 + sin q cos q) dq / (cos^2 q)

 = sec^2 q dq + sin q dq / cos q

Integrate to get:

ln r = tan q - ln(cos q) + C

r = A exp (tan q - ln(cos q))

 = A exp(tan q) / cos q

 = A exp (y/x) / (x/r)

x = A exp (y/x)

ln x = C + y/x

x ln x - C x = y

Check:
dy = (1+ln x - C) dx

(x+y) dx - x dy = (x + x ln x - C x) dx - x(1 + ln - C) dx = 0

---

Now that you've seen the method, you should be able to do the rest on your own.

-Doug Magnoli



Phishndisc wrote:

> I have these questions due and i cannot figure them out ANY help will be
> greatly appreciated...please email me with any any any! info....THANKS EVERYONE
>
> 1)  Suppose the equations M dx + N dy = 0 is homogenous.  Show that the
> transformation x = r cos q, y = r sin q reduces this equation to a separable
> equation in the variables r and q.   ( q= theta )
>
> 2)      Use this method above to solve these two -->
>
> ( x + y ) dx - x dy = 0
>
> and
>
> (2xy + 3 y^2) dx - ( 2xy + x^ 2) dy = 0
>
> 3)Solve The Diff. EQ.
>
> dy + (4y - 8 y^-3) x dx = 0
>
> 4) Solve initial value problems
>
> dy/dx + y = f(x),       Where f(x) = {2, 0 <= x <1
>                                                  {0, x >= 1,           y(0) = 0
>
> 5)  Solve The Diff. EQ.
>
> x dy/dx + (2x+1)/(x+1) y = x-1