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Tips on Solving Differential Equations

Date: 10/27/2003 at 07:19:52
From: David
Subject: Differential equations

I'm teaching myself differential equations with a great book 
(Differential Equations - With Applications and Historical Notes, 
George F. Simmons). Although the book is great, the questions often go 
beyond the methods discussed in the text.

I have reached a total impasse with the following questions, which I 
suspect means that I'm missing an entire (or several entire) 

xdy = (x^5 + x^3*y^2 + y)dx
(y + x)dy = (y - x)dx
xdy = (y + x^2 + 9y^2)dx

I've checked that they aren't exact and tried IF e^(Integral P dx)
and similar IF for y.

If you can help me I'll be very grateful !!

Date: 10/28/2003 at 11:18:48
From: Doctor Korsak
Subject: Re: Differential equations

Hello David,

Integrating factors can be difficult to discover, so there are various 
tricks you can try, like the exp(integral P(x)dx) factor.  That is a 
good one to use for linear differential equations of first order, but 
none of your three cases is linear.  

I would be curious to know what methods are covered up to where you 
are in the book for the above examples.  In my old college book, A 
First Course in Differential Equations by Rudolf E. Langer, some 
other tricks mentioned are:

Collecting terms into groups, and looking for something "recognizably 
transformable into a differential by some suitable multiplier".

Changes of variables: x = f(s,u)  y = g(s,u).

Equations with homogeneous coefficients, for which the substitution 

    x = s   y = su    or    x = su  y = u

    turns the equation into one with separable variables, eg. y^2dx +
    (x^2 - xy + y^2)dy = 0 becomes uds + (s^2 + 1)du = 0.

Equations with linear coefficients:
    (a1*x + b1*y + c1)dx + (a2*x + b2*y + c2)dy = 0

    Let s = x + h,  u = a2*x + b2*y + c2, then the equation becomes
    (A(s - h) + Bu + C)ds + udu = 0
    after collecting terms into expressions represented by A, B, C.
    If A = 0, the variables are separable; if A =/= 0, let -Ah + C =
    0, and that leads to an equation with homogeneous coefficients of
    degree 1.
Replacement by a pair of differential equations:
    Given P(x,y)dx + Q(x,y)dy = 0, let x and y be functions of t:
    dx/dt = Q(x,y)   dy/dt = -P(x,y), or using an integrating factor
    dx/dt = k(x,y)Q(x,y)   dy/dt = -k(x,y)P(x,y) if that helps to
    solve for x in terms of t and y or y in terms of x and t.
    Then elimination of t results in an equation relating x and y. 
If you are still having trouble I can help further--let me know.
- Doctor Korsak, The Math Forum 

Date: 10/28/2003 at 11:57:02
From: David
Subject: Thank you (Differential equations)

Dr. Korsak,

Thanks for the assistance - I'll try the methods given and see how I
get on. Also, thanks for leaving me with a challenge and not just
solving the problems.

Associated Topics:
College Calculus

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