Find the general solution to :

$t\frac{dy}{dt}+2y=e^{2(t-4)}$

$y(t)=\frac{e^{2*(t-4)}}{2*t}-\frac{e^{2*(t-4)}}{4*t^2}+\frac{C}{t}$

$2$

Find the unique solution that satisfies the initial condition $y(4)=0.$

$y(t)=\frac{e^{2*(t-4)}}{2*t}-\frac{e^{2*(t-4)}}{4*t^2}-\frac{9*C}{4*t}$

For this solution, $\frac{dy}{dt}(4)=\frac{1}{4}.$

Hint: Use the differential equation and the value of $y(4)$ to find $\frac{dy}{dt}(4).$