For the following initial value problem:
a)Find a formula for the solution.
b) State the domain of definition of the solution.
c) Describe what happens to the solution as it approaches the limit of its domain of definition. Why can't the solution be extended for more time?
I separated and integrated and got y(t)=sqrt(2ln|t-2|+C)-1 and I don't really know where to go from there.
Using the initial condition, C will be 2*ln|-2|
Differential Equations - Damon, Thursday, January 31, 2013 at 5:10pm
If y is to be zero when t = 0 then
1 = sqrt (2 ln 2 +C)
C = 1 -2 ln 2 = -.386
y = sqrt (2 ln|t-2| -.386) - 1
You can not take ln of zero or a negative number so t may not be equal to 2
Try some solutions around t = t to see what happens to y