Just use the definition of the Laplace-operator in 1-D you get that:
\begin{align*} u''(x) = k_c u(x) \end{align*} This equation has three possible solutions depending on the sign of $k_c$-
If $k_c>0$ the solution is: \begin{align*} u(x) = C_1e^{\sqrt{k_c}x}+C_2 e^{-\sqrt{k_c}x} \end{align*} Now inserting your initial value we have the constraint that $C_1+C_2 =1$.
If $k_c<0$ the solution is: \begin{align*} u(x) = C_1\sin({\sqrt{-k_c}x})+C_2 \cos({\sqrt{-k_c}x}) \end{align*} Again inserting your initial condition we have the constraint $C_2=1$.
If $k_c=0$ well that is left as an exercise ;o)