(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.
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.