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

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

Answer 2

The required solution is $y(6) = 20e^{-3}$

Integration

Given the following differential equation $dy/dt=-10e^{-t/2}$

To get the value of y, we will integrate the equation to have:

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

Given that y(0) = 20, hence;

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

Hence the solution to the given integral is $y=-20e^{-t/2}$

Next is to get y(6)

$y(6)=-20e^{-6/2} \\ y(6) =20e^{-3}}$

Hence the required solution is $y(6) = 20e^{-3}$

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