A copper rod 50 cm long with insulated lateral surface has initial temperature $u(x,0)=2x$, and a time $t=0$ its two ends are insulated. a.) Find $u(x,t)$. b.) What will its temperature be at $x=10$ after 1 min? c.) After approximately how long will its temperature at $x=10$ be 45 degrees celsius?
Here, $k=1.15 \frac{cm^2}{s}$ (according to my textbook). $L=50cm$, $u_x(0,t)=u_x(50,t)=0$. $f(x)=u(x,0)=2x$
$$a_0=\frac{2}{50}\int_0^{50}2xdx=100$$ $$a_n=\frac{2}{50}\int_0^{50}2xcos(\frac{n\pi x}{50})dx= 200\frac{((-1)^n-1)}{\pi^2n^2}$$
Finding $u(x,t)$ I get $$u(x,t)=50+\Sigma \hspace{1mm} 200\frac{((-1)^n-1)}{\pi^2n^2}e^{\frac{-1.15n^2\pi^2t}{50^2}}\cos(\frac{n^2\pi^2}{50^2}x)$$ where $n=1,2,3...$
Whenever I estimate b.), I get a number that is different from their answer. $u(10,60) \approx 19.159$, and the book gives the answer as $25.15$. Similarly for c.), the book gives the answer as 6 mins 54.3 seconds, whereas my answer yielded 7.68 minutes.
Am I simply miscalculating or am I misunderstanding some aspect of the problem? Any help is appreciated. Thank you.
Edit: I should also mention that I am estimating the series with $n=1$.