I've been asked to find solutions to the separable ODE given by $$(t^2-12t+27)\frac{dy}{dt}=y,\qquad y(6)=1.$$ and to find the interval on which the solution is valid. What I have so far is \begin{align*}
(t^2-12t+27)\frac{dy}{dt}&=y \\
\frac{1}{y}\frac{dy}{dt}&=\frac{1}{t^2-12t+27} \\
\int\frac{1}{y}dy&=\int\frac{1}{t^2-12t+27}dt \\
\log|y|&=\frac{1}{6}\log|t-9|-\frac{1}{6}\log|t-3|+C.
\end{align*} After using the initial value and a bit of calculation, I got that $C=0$. Carrying on, \begin{align*}
\log|y|&=\frac{1}{6}\log\left|\frac{t-9}{t-3}\right| \\
y&=\left(\frac{t-9}{t-3}\right)^{\frac{1}{6}}.
\end{align*} After doing all this, I got that the interval of validity should be $t\geq9$, but I got the answer incorrect after putting it into the system.
I just want to know where and why I went wrong. Even after keeping the absolute value, the interval of validity is for all $t\neq3$, and still got it wrong. Thank you.
Best Answer
Your works seems fine. The solution is required to exist around $t=6$, as that's where you have one of your conditions. So one reasonable answer would be $3<t<9$.