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#1 2005-10-30 04:47:54

Registered: 2005-10-30
Posts: 30

A differential equation problem.

3y'''+ 5y''-y' +7y=0

y(1)=0, y'(1)=0,y''(1)=0

ANS: y=0 is the only solution on any interval containing x=1



#2 2005-10-30 17:33:03

Registered: 2005-01-21
Posts: 7,555

Re: A differential equation problem.

Hmmm ...

you know that when x=1, the value of y=0, also the slope (y') is flat, and the change of slope (y'') is 0.

The only value you don't know at x=1 is y''', and that can be solved:

3y'''+ 5y''-y' +7y=0 ==> 3y'''+ 0 - 0 + 0=0 ==> y''' must equal 0

So, all given values are 0 and all flat at x=1 ...

Now what happens as you move away from x=1? If y were to change then the slope (y') will change, and you know that the slope is constrained by the equation 3y'''+ 5y''-y' +7y=0.

So let us say that as x increases, y increases. So the slope of y goes from 0 to positive, and so the rate of change in slope (y'') must also increases, and hence y''' also! So they must all increase together.

What would that do to: 3y'''+ 5y''-y' +7y=0 ?

I suspect (but haven't got as far as proving) that it is not possible to have y increase, because the various rates of change would make 3y'''+ 5y''-y' +7y 0.

Sorry, but that is as far as I have got.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman


#3 2005-10-31 01:59:05

Registered: 2005-10-30
Posts: 30

Re: A differential equation problem.



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