Answer
$y(x)=2e^{4x}-xe^{4x}$ is the solution to the initial-value problem
Work Step by Step
Solve the auxiliary equation for the differential equation. $$(r^2-8r+16)=0$$
Factor and solve for the roots. $$(r-4)^2=0$$
Roots are: $r_1=4$, as a multiplicity of $2$.
This implies that there are $\bf{Two}$ independent solutions to the differential equation and the general equation is equal to $y(x)=C_1 e^{4x}+C_2 x e^{4x}$
Now, the initial condition $y(0)=2$ yields $c_1=2$ and the initial condition $y^{\prime}(0)=7$ yields $4C_1+C_2=7 \implies (4)(2)+C_2=7$ or, $c_2=-1$
Therefore, $y(x)=2e^{4x}-xe^{4x}$ is the solution to the initial-value problem.