wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Differential Equation
        
(Message started by: THUDandBLUNDER on Dec 13^th, 2006, 4:38pm)

Title: Differential Equation
Post by THUDandBLUNDER on Dec 13^th, 2006, 4:38pm

Find y(x) such that y''(x) = y'(x)*y(x)

Title: Re: Differential Equation
Post by Icarus on Dec 13^th, 2006, 6:31pm

y=-2/x is one solution. (Go for the easy one first is my motto!)

Title: Re: Differential Equation
Post by Michael_Dagg on Dec 13^th, 2006, 7:49pm

A couple others are

y(x) = 0

and

y(x) = sqrt(2) tan(sqrt(2)/2 x)

In general

y(x) = sqrt(2) tan(sqrt(2)(x + c1)/(2c2) )/c2, for c2 =/= 0

are solutions as well.

Title: Re: Differential Equation
Post by Icarus on Dec 14^th, 2006, 6:10am

For a derivation: note that y'y = (1/2)(d/dx)y², so y" = (d/dx) (y²/2), and so y' = (y² + c2)/2 for some c2.

If c2 = 0, then y'/y² = 1/2, -1/y = x/2 + c1/2, or y = -2/(x + c1).

If c2 =/= 0, then y'/(y² + c2) = 1/2.

If c2 > 0, let s2 = sqrt(c2). Then Arctan (y/s2)/s2 = (x+c1)/2, or y = s2*tan(s2(x+c1)/2).

If c2 < 0, let s2 = sqrt(-c2). Then tanh^-1(-y/s2)/s2 = (x+c1)/2, or y = -s2*tanh(s2(x+c1)/2).

Title: Re: Differential Equation
Post by balakrishnan on Jan 6^th, 2007, 12:20pm

If we put
a=y''(x),v=y'(x)
the problem is same as
a=y*v
or
vdv/dy=y*v
which gives
v=y^2/2+C
or
dy/dx=(y^2+K)/2
which gives
y=C*tan(Cx/2+B)
where B,C are constants(including complex)

Sorry.I did not look at Icarus's post

Title: Re: Differential Equation
Post by Icarus on Jan 7^th, 2007, 3:27pm

It's alright. I'm glad you had the satisfaction of solving it yourself!