Answer
$y(x)=3 e^{2x}\cos x - e^{2x}\sin x$ is the solution to the initial-value problem.
Work Step by Step
Solve the auxiliary equation for the differential equation. $$(r^2-4r+5)=0$$
Roots are: $r_1=2-i; r?_2=2+i$
This implies that there are $\bf{Two}$ independent solutions to the differential equation and the general equation is equal to $y(x)=C_1 e^{2x}\cos x +C_2 e^{2x}\sin x$
Now, the initial condition $y(0)=3$ yields $c_1=3$ and the initial condition $y^{\prime}(0)=5$ yields $2C_1+C_2=5 \implies (2)(3)+C_2=5$ or, $C_2=-1$
Therefore, $y(x)=3 e^{2x}\cos x - e^{2x}\sin x$ is the solution to the initial-value problem.