0 like 0 dislike
0 like 0 dislike
Second order ODE with initial values at different times.

1 Answer

0 like 0 dislike
0 like 0 dislike
Are you sure the solution exist? I think you got a wrong equation.

The solution cannot have any point where y(x) and y'(x) are both nonzero and has the same sign. Assume there is such a point c>0, and since negation of a solution is still a solution, WLOG assume y(c),y'(c)>0. Then if there is a d>c such that y(d)<=0, then there is a y'(e)<=0 where c<e<d by mean value theorem; but if there are d>c such that y'(d)<0 then there is a y''(e)<0 where c<e<d but then y(e)<0, so there is some y(f)=0 for c<f<e. So there are no minimum d>c such that y(d)=0 or y'(d)=0, contradicting the fact that it must exist because y and y' are continuous.

Intuitively, if y'', y', and y all has the same sign, then the function keep being pushed in the same direction, so y and y' keep deviating further and further from 0.

No related questions found

33.4k questions

135k answers


33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!