Similar to how to solve $ {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$:
Let $f(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=ic^2X''(x)T(t)$
$\dfrac{T'(t)}{ic^2T(t)}=\dfrac{X''(x)}{X(x)}=-(K(k))^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-ic^2(K(k))^2\\X''(x)+(K(k))^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(k)e^{-ic^2t(K(k))^2}\\X(x)=\begin{cases}c_1(k)\sin((x-m)K(k))+c_2(k)\cos((x-m)K(k))&\text{when}~K(k)\neq0\\c_1x+c_2&\text{when}~K(k)=0\end{cases}\end{cases}$
$\therefore f(x,t)=C_1x+C_2+\int_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))~dk+\int_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))~dk$
or $C_1x+C_2+\sum\limits_kC_3(k)e^{-ic^2t(K(k))^2}\sin((x-m)K(k))+\sum\limits_kC_4(k)e^{-ic^2t(K(k))^2}\cos((x-m)K(k))$
The choice of integral or summation depends on number of boundary conditions. If there is only one boundary condition or no boundary conditions, we should choose integral. If there are two boundary conditions, we should choose summation.
Another brilliant method is called the power series method.
Similar to PDE - solution with power series:
Let $f(x,t)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^n}{n!}\dfrac{\partial^nu(a,t)}{\partial x^n}$ ,
Then $f(x,t)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(a,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(a,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{i^nc^{2n}(2n)!}\dfrac{\partial^nu(a,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}\dfrac{\partial^{n+1}(a,t)}{\partial t^n\partial x}=\sum\limits_{n=0}^\infty\dfrac{F^{(n)}(t)(x-a)^{2n}}{i^nc^{2n}(2n)!}+\sum\limits_{n=0}^\infty\dfrac{G^{(n)}(t)(x-a)^{2n+1}}{i^nc^{2n}(2n+1)!}=\sum\limits_{n=0}^\infty\dfrac{(-1)^nF^{(2n)}(t)(x-a)^{4n}}{c^{4n}(4n)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nF^{(2n+1)}(t)(x-a)^{4n+2}}{c^{4n+2}(4n+2)!}+\sum\limits_{n=0}^\infty\dfrac{(-1)^nG^{(2n)}(t)(x-a)^{4n+1}}{c^{4n}(4n+1)!}-\sum\limits_{n=0}^\infty\dfrac{i(-1)^nG^{(2n+1)}(t)(x-a)^{4n+3}}{c^{4n+2}(4n+3)!}$