Let $v(t,x)=\sum\limits_{k=1}^\infty a(k,t)\sin k\pi x+\sum\limits_{k=0}^\infty b(k,t)\cos k\pi x$ so that it automatically satisfies $v(t,x+2)=v(t,x)$ ,
Then $\sum\limits_{k=1}^\infty a_t(k,t)\sin k\pi x+\sum\limits_{k=0}^\infty b_t(k,t)\cos k\pi x=-\sum\limits_{k=1}^\infty k^2\pi^2a(k,t)\sin k\pi x-\sum\limits_{k=0}^\infty k^2\pi^2b(k,t)\cos k\pi x$
$\therefore a_t(k,t)=-k^2\pi^2a(k,t)$ and $b_t(k,t)=-k^2\pi^2b(k,t)$
$a(k,t)=A(k)e^{-k^2\pi^2t}$ and $b(k,t)=B(k)e^{-k^2\pi^2t}$
$\therefore v(t,x)=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x$
$v(t,x)=-v(t,-x)$
$\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x=-\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin(k\pi(-x))-\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos(k\pi(-x))$
$\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x-\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x$
$\sum\limits_{k=0}^\infty2B(k)e^{-k^2\pi^2t}\cos k\pi x=0$
$B(k)=0$
$\therefore v(t,x)=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x$
$v(0,x)=u_0(x)~\forall x\in[0,1]$ :
$\sum\limits_{k=1}^\infty A(k)\sin k\pi x=u_0(x)~\forall x\in[0,1]$
$A(k)=2\int_0^1u_0(x)\sin k\pi x~dx~\forall x\in[0,1]$
$\therefore v(t,x)=\sum\limits_{k=1}^\infty2\int_0^1u_0(x)\sin k\pi x~dx~e^{-k^2\pi^2t}\sin k\pi x~\forall x\in[0,1]$