Differential Equations (Another) Cont.
posted by Erica on .
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.
Differential Equations - Steve, Thursday, January 31, 2013 at 10:20am
y = √(2ln|t-2|+C)-1
we know that √x is defined only for x >= 0, so we must have
2ln|t-2| + C >= 0
ln|t-2| >= -C/2
t-2 >= e^(-C/2)
t >= 2+e^(-C/2)
In general, t>=2, but the form of the solution suggested that already.
As t -> 2, ln|t-2| -> infinity
Not sure why large t cannot be used. May be missing some of the characteristics of the problem.
The initial value conditons will determine C.
The initial condition was y(0)=0
I forgot to add it.
How do I solve for that?