$c_1=-1,-1$

Step $5$

We have that the general solution and the value of one of the constants is as follows.

$x(t)=-e^{-4t}{c}_{2}t{e}^{-4t}$

Taking the derivative of $x$ will allow us to solve for $c_2.$

$x^{\text{'}}(t)=A{e}^{-4t}-A{c}_{2}P{e}^{-4t}{c}_2()$

We are told that the mass is released with a downward velocity of $10f\frac{t}{s},$ or $x^{\text{'}}(0)=10.$ Substituting this value results in the following.

$$