Separate variables:

1/((t-18)(t-6))=1/y

Integrate, partial fraction decomposition (Checked with Wolfram alpha, these are correct):

1/(12(t-18))-1/(12(t-6))=1/y

ln((t-18)/(t-6))+c=12ln(y)

-----Now unsure

Raise everything to e and move the 12 over (Do I need to treat that 12 as an exponent after I raise everything to e?):

(t-18/(12(t-6))+c=y

Initial condition y(12)=1

thus c= 13/12

The are two other parts to this problem, asking what the range of t values the solution is valid for, which according to the variable-separated differential is 6 through 18. On there I am correct. The second part asks what the solutions approach as t approaches these limits, (infinity and 0) where I'm also correct.

So something I'm doing is wrong, but not fundamentally wrong enough to screw up the whole question.