Using Laplace transform method Using Laplace transform method, solve $$u_t=3u_{xx}$$
Subject to the boundary conditions
$u(π/2)=0 $ ,      $∂u/∂x (0,t)=0$ ,
And initial condition 
$u(x,0)= 30\cos{5x}$
 A: I think given the boundary conditions, you seek to solve $u_t=3 u_{xx}$; I will assume this in my solution below.
Laplace transform in $t$ and define
$$\hat{u}(x,s) = \int_0^{\infty} dt \, u(x,t) \, e^{-s t}$$
Then the equation is equivalent to
$$\frac{d^2}{dx^2} \hat{u}(x,s) - \frac13 s \hat{u}(x,s) = -u(x,0) = -10 \cos{5 x}$$
The easiest way to solve this is to assume a particular solution of $\hat{u}_p(x,s) = K \cos{5 x}$.  Substituting this into the equation, we get
$$-5^2 K - \frac{s}{3} K = -10 \implies K = \frac{30}{s+75}$$
The general solution is then
$$\hat{u}(x,s) = A e^{\sqrt{s/3} x} + B e^{-\sqrt{s/3} x} + \frac{30}{s+75} \cos{5 x} $$
Now apply the boundary conditions.
$$\frac{\partial u}{\partial x} (0,t)=0 \implies A-B=0$$
$$u\left ( \frac{\pi}{2},t\right) = 0 \implies A e^{(\pi/2)\sqrt{s/3}} + B e^{-(\pi/2)\sqrt{s/3}}=0$$
The above imply that $A=B=0$.  Thus
$$\hat{u}(x,s) = \frac{30}{s+75} \cos{5 x} $$
The solution we seek is then the inverse LT of $\hat{u}$, which is
$$u(x,t) = 30\, e^{-75 t} \, \cos{5 x} \:\theta(t)$$
where $\theta$ is the Heaviside (step) function.
