Question

If dy/dt=-10e^-t/2 and y(0)=20 what is the value of y(6)​

Answer 1

ANSWER

$y (6)= \frac{20}{ {e}^{3} }$

EXPLANATION

The given equation is:

$\frac{dy}{dt} = - 10 {e}^{ - \frac{t}{2} }$

We integrate with respect to t to get,

$y = 20 {e}^{ - \frac{t}{2} } + k$

The initial condition is y(0)=20

We apply the initial condition to get,

$20= 20 {e}^{ - \frac{0}{2} } + k$

$k = 0$

This implies that;

$y = 20 {e}^{ - \frac{t}{2} }$

$y (6)= 20 {e}^{ - \frac{6}{2} }$

$y (6)= 20 {e}^{-3}$

Or

$y (6)= \frac{20}{ {e}^{3} }$