= Solution
Substitute $u(y,t)=Y(y)T(t)$ into the <diffusion equation> and divide on a region where both factors are nonzero. This gives $Y''/Y=T'/(\nu T)$. The two sides depend on different variables, so they equal a separation <constant> $a$:
$$
\boxed{Y''=aY,\qquad T'=\nu aT.}
$$
These identities then extend through the zeros of the separated factors. The zero <Dirichlet boundary conditions> require $Y(0)=Y(1)=0$. For a nontrivial solution, $a$ must be negative: $a=0$ gives only a vanishing linear function and $a>0$ gives only a vanishing hyperbolic-sine solution. Writing $a=-k^2$ gives $Y=A\sin(ky)$, and $k=n\pi$, $n=1,2,\ldots$. Consequently the separated modes of <free viscous decay between parallel plates> are $\sin(n\pi y)e^{-\nu n^2\pi^2t}$.
The first initial condition selects the lowest mode:
$$
\boxed{u(y,t)=\sin(\pi y)e^{-\nu\pi^2t}.}
$$
The <superposition principle> for a homogeneous <linear differential equation> states that any linear combination of solutions is again a solution, with the same combination of homogeneous <boundary conditions> and initial data. It therefore gives
$$
\boxed{u(y,t)=\sin(\pi y)e^{-\nu\pi^2t}+\frac1{10}\sin(3\pi y)e^{-9\nu\pi^2t}.}
$$
The higher spatial mode decays nine times as fast, so the <velocity> profile approaches a multiple of the first sine mode before vanishing altogether.
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