Problem: If $~u(x,t)~$ is the solution of $$\dfrac{\partial u}{\partial t}=\dfrac{\partial^2 u}{\partial x^2},~~~~0<x<1,~~t>0\\u(x,0)=1+x+\sin(\pi x)\cos(\pi x)\\u(0,t)=1,~~u(1,t)=2$$then $$1.~~~u\left(\dfrac 12,\dfrac 14\right)=\dfrac 32~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.~~~u\left(\dfrac 14,\dfrac 34\right)=\dfrac 54+\dfrac 12e^{-3\pi^2}\\3.~~~u\left(\dfrac 12,\dfrac 12\right)=\dfrac 32~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~4.~~~u\left(\dfrac 14,1\right)=\dfrac 54+\dfrac 12e^{-4\pi^2}$$
My approach: We know that the general solution of the heat equation for one space variable $$\dfrac{\partial u}{\partial t}=\alpha\dfrac{\partial^2 u}{\partial x^2},~~~~0\le x\le L,~~t\ge0\\u(x,0)=f(x),~~~~\forall~~x\in[0,L]\\u(0,t)=0=u(L,t),~~~~\forall~~t>0$$is $$u(x,t)=\sum_{n=1}^\infty D_n \sin\left(\dfrac{n\pi x}{L}\right)e^{-n^2\pi^2\alpha t/L^2}$$where$$D_n=\dfrac 2L\int_0^L f(x)\sin\left(\dfrac{n\pi x}{L}\right) dx~.$$ Clearly here $~L=1,~\alpha=1~$ and $~f(x)=1+x+\sin(\pi x)\cos(\pi x).$ But one can see that the boundary values are different from the Heat equation. So I stuck here. Please help to solve the problem.
Here's a hint: consider $v(x,t)=u(x,t)-(1+x)$.