solution of diffusion equation The diffusion equation
$$\frac{\partial ^2 u}{\partial x^2} = \frac{\partial u}{\partial t},\;\;u = u(x,t),\;\;u(0,t) =0 = u( \pi ,t),\;\;u(x,0) = \cos x \sin 5x $$
admits the solution


*

*$( {e^{-36t}/2 )}  $ $ [\sin 6x + e^{20t} \sin 4x] $

*$( {e^{-36t}/2 )}  $ $ [\sin 4x + e^{20t} \sin 6x] $

*$( {e^{-20t}/2)}  $ $ [\sin 3x + e^{15t} \sin 5x] $

*$( {e^{-36t}/2 )}  $ $ [\sin 5x + e^{20t} \sin x] $


I am stuck on this problem. Can anyone help me please....
 A: You can eliminate choices 3) and 4) right away, because a) the diffusion equation is linear, and b) $\sin{5 x} \cos{x} = 1/2 ( \sin{6 x} + \sin{4 x} )$.
The solution $u(x,t) = A(t) \sin{6 x} + B(t) \sin{4 x}$.  You can find the equations for $A(t)$ and $B(t)$ independently.
A: You could go about this by simply calculating the general solution to this (quite common) 1D heat equation with Dirichlet boundary conditions at the left and right endpoints ($x=0$ and $x=\pi$), and then deducing the coefficients in the general solution from the initial condition.
However, on a multiple choice question, I'd tend to work a little more efficiently by doing elimination.
For example, we quickly see all four answers satisfy the boundary conditions $u(0,t)=0=u(\pi,t)$. (Do you see this?) So we move on to the initial condition. The identity $$\sin A\cos B={1\over 2}(\sin(A+B)+\sin(A-B)),$$ says that the given initial condition can be expressed as $$u(x,0)=\sin(5x)\cos x={1\over 2}(\sin(6x)+\sin(4x)).$$ So now we are down to either (1) or (2).
Finally, check (1) and (2) by plugging them into the PDE. You will find that (1) does indeed satisfy the PDE (because for (1), $u_{xx}-u_t=0$, just do the computation), but (2) does not (since for (2), $u_{xx}-u_t=10 e^{-36 t} \left(\sin (4 x)-e^{20 t} \sin (6 x)\right)$ rather than 0.)
Hence, the answer is (1).
