Help Solving Textbook Heat Conduction Laplace Transforms PDE Problem I am trying to solve the following problem:
$$\dfrac{\partial{\phi}}{\partial{t}} = \dfrac{\partial^2{\phi}}{\partial{x}^2} - \cos(x), \ x > 0, t > 0$$
$$\phi(x, 0) = 0, \ x > 0$$
$$\phi(0, t) = e^{-t}, \ t > 0$$
Taking the Laplace transform (in $t$, of course) gives
$$\mathcal{L} \left\{ \dfrac{\partial{\phi}}{\partial{t}} \right\} = \mathcal{L} \left\{ \dfrac{\partial^2{\phi}}{\partial{x}^2} - \cos(x) \right\}$$
$$\therefore s \mathcal{L}\{\phi\} - \phi(0) = \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2} - \dfrac{\cos(x)}{s}$$
And since $\phi(x, 0) = 0$, we have 
$$s \mathcal{L} \{ \phi \} = \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2} - \dfrac{\cos(x)}{s}$$
This is an ODE with constant coefficients. 
From now on, let $\mathcal{L}\{ \phi \} = \bar{\phi}$ (for the sake of conciseness). 
So our constant-coefficient ODE is
$$\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = \dfrac{\cos(x)}{s}$$
This is a inhomogeneous ODE and, judging by the inhomogeneous term $\dfrac{\cos(x)}{s}$, can be solved using the method of undetermined coefficients.
The homogeneous version of the ODE is 
$$\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = 0$$
The characteristic polynomial is 
$$m^2 - s = 0$$
$$\therefore m = \pm \sqrt{s}$$
Therefore, the complementary equation for this inhomogeneous ODE is
$$y_c(x) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}}$$
And our particular solution for this inhomogeneous ODE is 
$$y_p(x) = -\dfrac{1}{s^2 - s} \cos(x)$$
Therefore, our solution to the inhomogeneous ODE is
$$y(x) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}} - \dfrac{1}{s^2 - s} \cos(x)$$
Therefore, we have that 
$$\bar{\phi}(x, s) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}} - \dfrac{1}{s^2 - s} \cos(x)$$
Here is where I am stumped. 
HOWEVER, one way that I have seen other problems proceed is by assuming another boundary condition: 
$$\lim\limits_{x \to \infty} \phi(x, t) = 0$$
This makes sense from a physical standpoint, but it was not specified in the problem statement, so I am unsure if it is valid for me to use it? 
Also, I don't have full solutions for these problems, so I have no way of checking the intermediate steps.
The final solution should be
$$\phi(x, t) = \text{erfc}\left( \dfrac{x}{2\sqrt{t}} \right) - \cos(1 - e^{-t})$$
I would greatly appreciate it if people could please help me solve this problem.

EDIT: If we proceed with assuming the extra boundary condition (that is, if we proceed with assuming that $\lim\limits_{x \to \infty} \phi(x, t) = 0$), then we continue from above as follows:
Now, since $\lim\limits_{x \to \infty} \phi(x, t) = 0$, we have that $Ae^{-x \sqrt{s}} = \dfrac{A}{e^{x \sqrt{s}}} \to 0$ as $x \to \infty$, and $- \dfrac{1}{s^2 - s} \cos(x)$ does not exist as $x \to \infty$, since $\cos(x)$ will not converge.
So what do we do from here? It seems that the $- \dfrac{1}{s^2 - s} \cos(x)$ term is causing us problems?
IGNORE EVERYTHING BELOW THIS POINT

$$\bar{\phi}(x, s) = \int_0^\infty \phi(x, t) e^{-st} \ dt,$$
we have that $\bar{\phi}(x, s) \to 0$ as $x \to \infty$, and hence $B = 0$, since $Ae^{-x \sqrt{s}} = \dfrac{A}{e^{x \sqrt{s}}} \to 0$ as $x \to \infty$.
Therefore, we now have
$$\bar{\phi}(x, s) = Ae^{-x \sqrt{s}}.$$
We now use the given boundary conditions.
Letting $x = 0$ in the Laplace transform of $\phi$ gives
$$\bar{\phi}(0, s) = \int_0^\infty \phi(0, t)e^{-st} \ dt = \int_0^\infty e^{-t}e^{-st} \ dt = \int_0^\infty e^{-t(1 + s)} \ dt$$
$$= \dfrac{-1}{1 + s} \int_0^{-\infty} e^u \ du = \dfrac{1}{1 + s},$$
since $\phi(0, t) = e^{-t}$ for all $t > 0$.
 A: Let $\widetilde{\phi}(x,s) = \int_{0}^{\infty}e^{-ts}\phi(x,t)dt$ be the Laplace transform of $\phi(x,t)$. As you said we have the following ODE for $\widetilde{\phi}(x,s)$:
$$
\dfrac{\partial^2{\widetilde{\phi}(x,s)}}{\partial{x}^2} -s\widetilde{\phi}(x,s)- \frac{\cos(x)}{s} = 0
$$
The general solution is simply:
$$
\widetilde{\phi}(x,s)=Ae^{\sqrt{s}x}+Be^{-\sqrt{s}x} -\frac{\cos(x)}{s(s+1)}
$$
where $A$ and $B$ are two arbitrary constant that need two condition to be determined. The first condition is a physical one: the solution must be limited in time and so $A=0$.
The second condition is that:
$$\phi(0, t) = e^{-t}, \ t > 0 .$$
Now the Laplace transform of $\phi(0,t)$ is simply $\widetilde{\phi}(0,s)=\frac{1}{1+s}$ and so we have $B = \frac{1}{s}$. Finally we have that the Laplace transform of our $\phi(x,s)$ is:
$$
\widetilde{\phi}(x,s)=\frac{e^{-\sqrt{s}x}}{s} - \frac{\cos(x)}{s(s+1)}
$$
Now we have to transforming back. I don't know if you know complex analysis. If not, simply check a Laplace transform's table and see that the inverse Laplace transform of $\frac{e^{-\sqrt{s}x}}{s}$ is $\text{erfc}(\frac{x}{2\sqrt{t}})$ and the inverse L.T. of $\frac{1}{s(s+1)}$ is $1 - e^{-t}$ and so the solution is:
$$ \phi(x,t) = \text{erfc}\left( \dfrac{x}{2\sqrt{t}} \right) - \cos(x)(1 - e^{-t}).
$$
If you know complex analysis you have to perform the following complex integral:
$$
\phi(x,t) = \frac{1}{2{\pi}i}\int_Ldze^{zt}\widetilde{\phi}(x,z)
$$
where L is a vertical line in the complex plane on the right of all singularity of $\widetilde{\phi}(x,z)$. Be careful due to the presence of the square root: is a multivalued function. 
A: Let $\Phi(s,x)=\int_{0}^{\infty}\phi(t,x)e^{-st}dt$ be the Laplace transform in time of the desired solution. Then the transform of the heat equation is
$$
    \int_{0}^{\infty}e^{-st}\frac{\partial\phi}{\partial t}dt
  =\int_{0}^{\infty}e^{-st}\frac{\partial^2\phi}{\partial x^2}dt-\cos(x)\int_{0}^{\infty}e^{-st}dt.
$$
Expanding this out gives
$$
   e^{-st}\phi(t,x)|_{t=0}^{\infty}-\int_{0}^{\infty}(-se^{-st})\phi(t,x)dt \\
  =\frac{\partial^2}{\partial x^2}\int_{0}^{\infty}e^{-st}\phi(t,x)dt-\left.\cos(x)\frac{e^{-st}}{-s}\right|_{t=0}^{\infty}.
$$
Therefore,
$$
   -\phi(0,x)+s\int_{0}^{\infty}e^{-st}\phi(t,x)dt=\frac{\partial^2}{\partial x^2}\int_{0}^{\infty}e^{-st}\phi(t,x)dt-\frac{\cos(x)}{s}.
$$
Let $\Phi(s,x)=\int_{0}^{\infty}e^{-st}\phi(t,x)dt$. Then $\phi(0,x)=0$ gives
$$
     s\Phi(s,x)=\Phi_{xx}(s,x)-\frac{\cos(x)}{s} \\
      \Phi_{xx}-s\Phi=\frac{\cos(x)}{s}.
$$
A particular solution is $C\cos(x)$. But $\cos''=-\cos$, which gives $C(-1-s)=1/s$ or $C=-1/(s(1+s))$. So the general solution is
$$
      \Phi(s,x)=A\cos(\sqrt{s}x)+B\sin(\sqrt{s}x)-\frac{\cos(x)}{s(1+s)}.
$$
Now you have to transform back in $s$.
