(t\(^{2}\)-24t+108)dy/dt=y

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.

 

Oh wow, latex did not work for me. Editing.

 

What happen to the constant c when you exponentiate everything?

e.g.

y = ln(x) + c

what is e^y?

To see what happens to the 12, consider a simpler case in which you have

y = 12x

what is e^y?

Now change that to

y = 12ln(x)

what is e^y