Answer
$y(x)=e^{mx}\sin (kx) $
Work Step by Step
Solve the auxiliary equation for the differential equation. $$r^2-2mr+(m^2+k^2)=0$$
Roots are: $ r_1=m-ki; r_2=m+ki$
This implies that there are $\bf{3}$ independent solutions to the differential equation and the general equation is equal to $y(x)=C_1 e^{mx}\cos (kx) +C_2 e^{mx} \sin (kx)$
Next, after applying the initial conditions we get: $C_1=-0; C_2=1$
Therefore, $y(x)=e^{mx}\sin (kx) $ is the solution to the initial-value problem.