Pde using laplace transform Could you help me to find a solution for this partial differntial equation by using laplace transform 
$$u_{t} - u_{xx} = xt$$
where $$u(0,t)=t, \quad u(1,t)=t^2, \quad u(x,0)= \sin \pi x$$
I tried and I found $u(x,s)$ 
$$u(x,s) = c_{1} \exp \bigg( \frac{\sqrt{s}}{2} \bigg) + c_{2} \exp \bigg( - \frac{\sqrt{s}}{2} \bigg) + \frac{x}{s^3} + \frac{\sin( \pi x)}{4 \pi^2-s}$$
but I got stuck with the laplace transform.
 A: Let me start from scratch.  First we LT in time to get an ODE which is a boundary value problem.  We solve the ODE using standard techniques and solve for unknown coefficients using the boundary conditions.  Then we will speak of inverting the LT.
Define the LT solution as
$$U(x,s) = \int_0^{\infty} dt \, u(x,t) e^{-s t} $$
Then the PDE in $t$ may be rewritten as the ODE in $s$
$$\frac{d^2}{d x^2} U(x,s) - s U(x,s) = -\frac{x}{s^2} - \sin{\pi x} $$
$$U(0,s) = \frac1{s^2} \quad U(1,s) = \frac{2}{s^3}$$
The homogeneous solution to the above equation is
$$U_H(x,s) = A e^{\sqrt{s} x} + B e^{-\sqrt{s} x} $$
The inhomogeoeus part, $U_I(x,s)$, may be found by finding constants $C$ and $D$ such that 
$$U_I(x,s) = C x + D \sin{\pi x}$$
satisfies the ODE.  Thus, we have
$$-\pi^2 D \sin{\pi x} - s C x - s D \sin{\pi x} = -\frac{x}{s^2} - \sin{\pi x} $$
or
$$C = \frac1{s^3} \quad D = \frac1{s+\pi^2}$$
Thus,
$$U(x,s) = A e^{\sqrt{s} x} + B e^{-\sqrt{s} x} + \frac{x}{s^3} + \frac{\sin{\pi x}}{s+\pi^2}$$
Before we can invert the LT, we must find the unknown $A$ and $B$, as surely they depend on $s$.  We use the boundary conditions above:
$$U(0,s) = A + B = \frac1{s^2} $$
$$U(1,s) = A e^{\sqrt{s}} + B e^{-\sqrt{s}} + \frac1{s^3} = \frac{2}{s^3} $$
or
$$A + B = \frac1{s^2} $$
$$A e^{\sqrt{s}} + B e^{-\sqrt{s}} = \frac{1}{s^3} $$
so that
$$A = \frac{1-s e^{-\sqrt{s}}}{2 s^3 \sinh{\sqrt{s}}} $$
$$B = \frac{s e^{\sqrt{s}}-1}{2 s^3 \sinh{\sqrt{s}}} $$
We can rearrange the furniture a bit so we get the following nice form for $U$:
$$U(x,s) = \frac{\sinh{\left (\sqrt{s} x\right )}}{s^3 \sinh{\sqrt{s}}} +\frac{\sinh{\left [\sqrt{s} (1-x)\right ]}}{s^2 \sinh{\sqrt{s}}} + \frac{x}{s^3} + \frac{\sin{\pi x}}{s+\pi^2}$$
The good news is that, despite the square roots, there are no branch points here.  That means we can use a simple contour of integration and apply the residue theorem to find the inverse LT.  The bad news is that the ultimate answer will be an infinite series with no simple closed form as far as I know.  
The solution we seek is given by the ILT
$$u(x,t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, U(x,s) e^{s t} $$
which we may close off using a circular arc to the left of the line $\operatorname{Re}{z} = c$, i.e., $t \gt 0$.  $u(x,t)$ is then the sum of the residues at the poles $z=-n^2 \pi^2$, $n \in \{0, 1, 2, 3, \ldots \}$.  At $n=0$, we have two terms with poles of order $3$ and one with a pole of order $2$.  At $n=1$, we have three terms with simple poles.  For the other values of $n$, we have two terms with simple poles.
To get the double and triple poles, we expand the hyperbolic sines and the exponential term into Laurent expansions.  I will spare you the details for now, but the result is
$$\operatorname*{Res}_{z=0}\left [ U(x,z) e^{z t} \right ] = x t^2 + \frac16 (x^3-7 x+6) t + \frac1{360} (3 x^5 - 70 x^3 + 180 x^2 - 113 x) $$
The rest of it is straightforward because of the simple poles.  The result is

$$u(x,t) = x t^2 + \frac16 (x^3-7 x+6) t + \frac1{360} (3 x^5 - 70 x^3 + 180 x^2 - 113 x) \\+\sum_{n=1}^{\infty} a_n \sin{(n \pi x)} \, e^{-n^2 \pi^2 t} $$

where

$$a_1 = -\frac{2 }{\pi^5} - \frac{2 }{\pi^3}+ \frac{1}{\pi^2}$$

and for $n \ge 2$,

$$a_n = \frac{2 (-1)^n}{n^5 \pi^5} - \frac{2 }{n^3 \pi^3} $$

