Find the longest interval in which the solution of the initial value problem
(t^2 - 3t)y'' + ty' - (t + 3)y = 0, y(1) = 2, y'(1) = 1
is certain to exist.
If the given differential equation is written in the form of y'' + p(t)y' + q(t)y = g(t), y(t_0) = y_0, y'(t_0) = y'_0 then p(t) = 1/(t - 3), q(t) = -(t + 3)/t(t - 3) and g(t) = 0. The only points of discontinuity of the coefficients are t = 0 and t = 3. Therefore, the longest open interval, containing the initial point t = 1, in which all the coefficients are continuous is 0 < t < 3. Thus this is the longest interval in which the theorem guarantees that the solution exists.
Taken from an example from Boyce & DiPrima's book on differential equations.
Why is it that p(t) = 1/(t - 3)? Why is not p(t) = t?
August 14th 2012, 01:37 PM
Re: Longest interval?
Your generic form, y''+ p(t)y'+ q(t)y= g(t) has 1 as the coefficient of y''. Your given equation does not: the coefficient of y'' is t^2- 3t. To get it to the correct form, you need to divide through by t^2- 3t:
y''+ [t/(t^2+ 3t)] y'- [(t+ 3)/(t^2+ 3t)]y= 0. t^2+ 3t= t(t+ 3) so after reducing the fractions, that is y''+ [1/(t+3)]y'- [1/t]y= 0. p(t)= 1/(t+3) and q(t)= -1/t.