Initial Value Problem for an Inhomogeneous Equation Find the solution of this initial value problem. 
$u_t = 2u_{xx} + 6u$, $0<x<\pi$, $t>0$ 
$u(0,t) = 0 = u(\pi,t)$, $t>0$ 
$u(x,0)=5\sin3x-2\sin4x+3\sin10x$. 
Can someone help get me started?
 A: Let us use separation of variables. This means we assume that $u(x,t)$ takes the form
$$
u(x,t) = X(x)T(t)
$$
Plugging in to the original equation gives:
$$
X(x)T'(t) = 2X''(x)T(t) + 6X(x)T(t)
$$
Dividing through by $X(x)T(t)$:
$$
\frac{T'(t)}{T(t)} = 2\frac{X''(x)}{X(x)} + 6
$$
The left hand side is dependent only on $t$ and the right hand side is dependent only on $x$, yet they equal each other for all $(x,t)$. The only way this can be true is if each side is constant:
$$
2\frac{X''(x)}{X(x)} + 6 = -\lambda
$$
$$
\frac{T'(t)}{T(t)} = -\lambda
$$
Reordering the first equation gives
$$
X''(x) + \frac{\lambda+6}{2}X(x) = 0
$$
If we let $k^2=\frac{\lambda+6}{2}$, the general solution to this equation is
$$
X(x) = c_1\sin(kx)+c_2\cos(kx)
$$
Now because $u(0,t)=u(\pi,t)=0$, this translates to $X(0)=X(\pi)=0$. This is only possible if the $\cos$ term vanishes and $k$ takes on certain discrete values to match those boundary conditions. We get a family of solutions
$$
X_n(x) = c_n\sin(k_n x),\;\;k_n=n
$$
Now we return to the time equation, this time knowing that the values of the unknown constant are restricted to discrete values $\lambda_n=2k_n^2-6$; reordering that gives
$$
T_n'(t) + \lambda_n T_n(t) = 0
$$
With solution family
$$
T_n(t) = e^{-\lambda_n t}
$$
Now $u$ has a solution family
$$
u_n(x,t) = X_n(x)T_n(t) = c_n\sin(k_n x)e^{-\lambda_n t}
$$
That is a solution for any integer value of $n$, so the most general solution is a superposition of those solutions
$$
u(x,t) = \sum_{n=-\infty}^\infty u_n(x,t) =
\sum_{n=-\infty}^\infty c_n\sin(k_n x)e^{-\lambda_n t}
$$
To determine the unknown $c_n$ coefficients, we have to apply the initial condition
$$
5\sin(3x)-2\sin(4x)+3\sin(10x) = u(x,0) = \sum_{n=-\infty}^\infty c_n\sin(k_n x)
$$
Recalling that $k_n=n$, it is clear that the sequence of $c_n$ values is mostly zero, except that $c_3=5,\;c_4=-2,\;c_{10}=3$. So the final solution is 
$$
u(x,t) = 5\sin(3 x)e^{-12 t}-2\sin(4 x)e^{-26 t}+3\sin(10 x)e^{-194 t}
$$
A: $\newcommand{\+}{^{\dagger}}%
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$$
{\rm u}\pars{x,t}=
{\rm A}_{3}\pars{t}\sin\pars{3x} - {\rm A}_{4}\pars{t}\sin\pars{4x}
+ {\rm A}_{10}\pars{t}\sin\pars{10x}
$$

\begin{align}
&\dot{{\rm A}}_{3}\pars{t}\sin\pars{3x} - \dot{{\rm A}}_{4}\pars{t}\sin\pars{4x}
+ \dot{{\rm A}}_{10}\pars{t}\sin\pars{10x}
\\[3mm]&=
\bracks{-18{\rm A}_{3}\pars{t}\sin\pars{3x} + 32{\rm A}_{4}\pars{t}\sin\pars{4x}
-200 {\rm A}_{10}\pars{t}\sin\pars{10x}}
\\[3mm]&\phantom{=}+
\bracks{6{\rm A}_{3}\pars{t}\sin\pars{3x} - 6{\rm A}_{4}\pars{t}\sin\pars{4x}
+ 6{\rm A}_{10}\pars{t}\sin\pars{10x}}
\end{align}

$$\left\lbrace%
\begin{array}{rcrclrcr}
\dot{\rm A}_{3}\pars{t} & + & 12\,{\rm A}_{3}\pars{t} & = & 0\,,\qquad
&{\rm A}_{3}\pars{0} & = & 5
\\[1mm]
\dot{\rm A}_{4}\pars{t} & + & 26\,{\rm A}_{4}\pars{t} & = & 0\,,\qquad
&{\rm A}_{4}\pars{0} & = & -2
\\[1mm]
\dot{\rm A}_{3}\pars{t} & + & 194\,{\rm A}_{3}\pars{t} & = & 0\,,\qquad
&{\rm A}_{10}\pars{0} & = & 3
\end{array}\right.
$$

$$
{\rm A}_{3}\pars{t} = 4\expo{-12t}\,,\qquad
{\rm A}_{4}\pars{t} = -2\expo{-26t}\,,\qquad
{\rm A}_{3}\pars{t} = 3\expo{-194t}
$$

$$
\color{#00f}{\large{\rm u}\pars{x,t}=
4\sin\pars{3x}\expo{-12t} + 2\sin\pars{4x}\expo{-26t}
+ 3\sin\pars{10x}\expo{-194t}}
$$
