How do I get Mathematica to solve the following partial differential equation

$$ \frac{\partial u(y,t)}{\partial t} = \nu \frac{\partial^2 u(y,t)}{\partial y^2}+\frac{\partial U_0(t)}{\partial t}$$

where $U_0 = U_{0m} \cdot sech(\frac{2 \pi}{T} (t-T))$, $(T,U_{0m})$=constant

with the following boundary conditions:

$$ u=0 \; {\rm for} \; y = 0 $$ $$ u=U_0 \; {\rm for} \;y \rightarrow \infty$$

Have tried

U0[t_] := U0m Sech[2 \[Pi]/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];
DSolve[
    {eq, u[0, t] == 0, limit[u[y, t], y -> Infinity] == U0[t]},
    u[y, t],
    {y, t}
]

How do I get Mathematica to solve the following partial differential equation

$$ \frac{\partial u(y,t)}{\partial t} = \nu \frac{\partial^2 u(y,t)}{\partial y^2}+\frac{\partial U_0(t)}{\partial t}$$

where $U_0 = U_{0m} \cdot sech(\frac{2 \pi}{T} (t-T))$, $(T,U_{0m})$=constant

with the following boundary conditions:

$$ u=0 \; {\rm for} \; y = 0 $$ $$ u=U_0 \; {\rm for} \;y \rightarrow \infty$$

Have tried

U0[t_] := U0m Sech[2 π/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];
DSolve[
    {eq, u[0, t] == 0, limit[u[y, t], y -> Infinity] == U0[t]},
    u[y, t],
    {y, t}
]

How do I get mathematica to solve the following partial differential equation

$$ \frac{\partial u(y,t)}{\partial t} = \nu \frac{\partial^2 u(y,t)}{\partial y^2}+\frac{\partial U_0(t)}{\partial t}$$

where $U_0 = U_{0m} \cdot sech(\frac{2 \pi}{T} (t-T))$, $ (T,U_{0m})$=constant

with the following boundary conditions $$ u=0 \; for \; y = 0 $$ $$ u=U_0 \; for \;y \rightarrow \infty$$

Have tried

U0[t_] := U0m Sech[2 \[Pi]/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];
DSolve[
    {eq, u[0, t] == 0, limit[u[y, t], y -> Infinity] == U0[t]},
    u[y, t],
    {y, t}
]

How do I get Mathematica to solve the following partial differential equation

$$ \frac{\partial u(y,t)}{\partial t} = \nu \frac{\partial^2 u(y,t)}{\partial y^2}+\frac{\partial U_0(t)}{\partial t}$$

where $U_0 = U_{0m} \cdot sech(\frac{2 \pi}{T} (t-T))$, $(T,U_{0m})$=constant

with the following boundary conditions:

$$ u=0 \; {\rm for} \; y = 0 $$ $$ u=U_0 \; {\rm for} \;y \rightarrow \infty$$

Have tried

U0[t_] := U0m Sech[2 \[Pi]/T (t - T)];
eq = D[u[y, t], t] == nu D[D[u[y, t], y], y] - D[U0[t], t];
DSolve[
    {eq, u[0, t] == 0, limit[u[y, t], y -> Infinity] == U0[t]},
    u[y, t],
    {y, t}
]