What are the conditions for the Initial Condition Problem Theorem to hold true?

[the_kool_guy]

theorem states that for y'' + p(x)y' + q(x)y = r(x);
if y1 & y2 are random numbers such that y(m) = y1 and y'(m) = y2 then we can find a unique solution y for above differential equation...


in y'' - y'/x = 0;
both y = x^2 and y = 0 and y = k x^2 satisfy above with ..
y(0) = 0; and y'(0) = 0



then isn't this contradicts above theorem ,,
i have read that theorem is always true...

can someone enlighten me on this...?

thanks

 

[Mark44]

theorem states that for y'' + p(x)y' + q(x)y = r(x);
if y1 & y2 are random numbers such that y(m) = y1 and y'(m) = y2 then we can find a unique solution y for above differential equation...


in y'' - y'/x = 0;
both y = x^2 and y = 0 and y = k x^2 satisfy above with ..
y(0) = 0; and y'(0) = 0



then isn't this contradicts above theorem ,,
i have read that theorem is always true...

Aren't you omitting some of the theorem you're citing? For instance, what are the conditions on the p(x) and q(x) functions?

can someone enlighten me on this...?

thanks

 

[the_kool_guy]

hmmm ... p and q must be continuous over [a,b] where m lies in the above interval...

1/x is not continuous ... on 0
thanks

 

[HallsofIvy]

hmmm ... p and q must be continuous over [a,b] where m lies in the above interval...

1/x is not continuous ... on 0
thanks

I believe you will find that p' must also be continuous on (a, b).