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

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kudzordzifrancis kudzordzifrancis · Answer 1 · 2019-05-10T06:49:51+02:00

ANSWER

[tex]y (6)= \frac{20}{ {e}^{3} } [/tex]

EXPLANATION

The given equation is:

[tex] \frac{dy}{dt} = - 10 {e}^{ - \frac{t}{2} } [/tex]

We integrate with respect to t to get,

[tex]y = 20 {e}^{ - \frac{t}{2} } + k[/tex]

The initial condition is y(0)=20

We apply the initial condition to get,

[tex]20= 20 {e}^{ - \frac{0}{2} } + k[/tex]

[tex]k = 0[/tex]

This implies that;

[tex]y = 20 {e}^{ - \frac{t}{2} } [/tex]

[tex]y (6)= 20 {e}^{ - \frac{6}{2} } [/tex]

[tex]y (6)= 20 {e}^{-3} [/tex]

Or

[tex]y (6)= \frac{20}{ {e}^{3} } [/tex]

abidemiokin abidemiokin · Answer 2 · 2022-01-24T13:11:34+01:00

The required solution is [tex]y(6) = 20e^{-3}[/tex]

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

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

[tex]y=-20e^{-t/2} + k[/tex]

Given that y(0) = 20, hence;

[tex]20=-20e^{-0/2} + k\\ 20 = 20 + k\\ k = 0[/tex]

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

Next is to get y(6)

[tex]y(6)=-20e^{-6/2} \\ y(6) =20e^{-3}}[/tex]

Hence the required solution is [tex]y(6) = 20e^{-3}[/tex]

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