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        riddles >> putnam exam (pure math) >> differential equations
        
(Message started by: Nikki on May 26^th, 2004, 3:58am)

Title: differential equations
Post by Nikki on May 26^th, 2004, 3:58am

Consider the differential equation

2x.(dy/dx)+y3 = 0 x>0

a) Find all the solutions of this differential equation, carefully stating the largest open interval on which each particular solution is valid

b) Find the solution that satisfies y(1) = -1. On what interval is the solution valid?

Title: Re: differential equations
Post by THUDandBLUNDER on May 26^th, 2004, 4:26am

Yes, I'm sure somebody will be happy to do all your homework for you. ::)

Title: Re: differential equations
Post by nikki on May 26^th, 2004, 7:27am

never mind,i think this is the answer

dy/dx + y3 = 0

dy/dx = -y3

dy/-y3 = dx

1/(2y2) = x + C

2y2 = 1/(x + C)

y2 = 1/[2(x + C)]

y = ± sqrt(1/[2(x + C)])

1/[2(x + C)] > 0

2(x + C) > 0

x+C > 0

x > -C and x > 0

1/(2*12) = -1 + C

1/2 = -1 + C

3/2 = C

1/(2y2) = x + 3/2

1/(2y2) = (2x + 3)/2

2y2 = 2/(2x + 3)

y2 = 1/(2x + 3)

y = ± sqrt(1/(2x + 3))

2x + 3 > 0

2x > -3

x > -3/2

But we have the condition that x > 0.

Hence x > 0 it our interval

Title: Re: differential equations
Post by THUDandBLUNDER on May 26^th, 2004, 8:57am

Nikki, now that you have shown your working, I am even more sure you will get lots of help,
especially as Putnams are evidently getting rather easy these days.

;)

(Instead of writing 'x2', you can write x² by using the sup button above the edit window.)

Title: Re: differential equations
Post by Sir Col on May 26^th, 2004, 10:13am

on 05/26/04 at 03:58:00, Nikki wrote:

Consider the differential equation

2x.(dy/dx)+y3 = 0 x>0

Nikki, you seem to have lost 2x in your solution, but you had the right idea. From your differential equation...

2x dy/dx = -y³
1/(2x) dx/dy = -1/y³
[int] 1/(2x) dx = [int] -1/y³ dy
(1/2)ln(2x) = 1/(2y²)+c

Multiply through by 2 and replace constant with -ln(k)

ln(2x) = 1/y²-ln(k)
ln(2x)+ln(k) = 1/y²
ln(2kx) = 1/y²
y = [pm]1/[sqrt]ln(2kx)

OR

2kx = exp(1/y²)
x = exp(1/y²)/(2k)

I'll leave you to figure the rest out. ;)