For the following initial value problem:

dy/dt=1/((y+1)(t-2))
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.

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The initial condition was y(0)=0
I forgot to add it.
Then C=2*ln|-2|.
How do I solve for that?

To solve for C, you need to use the initial condition y(0) = 0 in the formula for the solution.

Substituting t = 0 and y = 0 into the equation y(t) = sqrt(2ln|t-2|+C)-1, we get:

0 = sqrt(2ln|0-2|+C)-1

Simplifying, we have:

0 = sqrt(2ln|(-2)|+C)-1

Taking the square of both sides, we get:

0 = 2ln|-2|+C-1

Since ln|-2| is undefined, we cannot directly solve for C using the initial condition y(0) = 0. This means that the solution may not be valid for t = 0 and the domain of definition of the solution may be restricted.

To further analyze the solution and determine the value of C, it would be helpful to have more information or additional initial conditions.