If $u(x,t)$ is the solution to $$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^2}~, \quad 0<x<1,~t>0 \\ u(x,0) = 1 + x + \sin(\pi x) \cos(\pi x)~, \quad u(0,t)=1,~u(1,t)=2$$ then
1).$u(\frac{1}{2},\frac{1}{4})=\frac{3}{2}$
2).$u(\frac{1}{2},\frac{1}{2})=\frac{3}{2}$
3).$u(\frac{1}{4},\frac{3}{4})=\frac{5}{4}+\frac{1}{2}e^{-3\pi^{2}}$
4).$u(\frac{1}{4},1)=\frac{5}{4}+\frac{1}{2}e^{-4\pi^{2}}$
I tried to solve this question: we know that solution of such equation is of the type: $$( c_1 \cos\lambda x + c_2 \sin\lambda x )\cdot c_3 e^{-\lambda^2 t}$$
but how can we get that $x$ in the solution equation?
It seems very complicated. Please help.
First, find the steady-state solution $w(x)$, which satisfies
$$ \frac{d^2 w}{dx^2} = 0, \quad w(1) = 1, w(2) = 2 $$
this gives $w(x) = 1 + x$
Next, we have
$$ u(x,t) = 1 + x + v(x,t) $$
where $v(x,t)$ also solves the heat equation, but is homogeneous on the boundary, i.e. $v(0,t)=v(1,t)=0$. Performing separation of variables on $v(x,t)$ gives the general solution
$$ u(x,t) = 1 + x + \sum_{n=1}^\infty a_n\sin(n\pi x)e^{-n^2\pi^2 t} $$
For a multiple choice question, you need to get this quick, so it may be helpful to memorize that Dirichlet B.C.s = sine series in $x$.
The initial condition is
$$ u(x,0) = 1 + x + \sum_{n=1}^\infty a_n\sin(n\pi x) = 1 + x + \frac12 \sin(2\pi x) $$
which you can compare terms to get $a_2 = \frac12$, and $a_n=0$ for $n\ne 2$.
Altogether, the final solution is
$$ u(x,t) = 1 + x + \frac12 \sin(2\pi x)e^{-4\pi^2 t} $$
which gives
$$ u\left(\frac12, \frac14\right) = \frac32 $$