3y'''+ 5y''-y' +7y=0
ANS: y=0 is the only solution on any interval containing x=1
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