The differential equation below models the temperature of a 95°C cup of coffee in a 19°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 70°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. dy dt = − 1 51 (y − 19)

y (t) = 19 + 76*e^(-(1/51)*t)

Step-by-step explanation:

We have the following differential equation model:

dy / dt = - (1/51) * (y -19)

The initial conditions are: y (0) = 95 and y '(70) = 1

We solve the differential equation, separating variables:

1 / (and -19) * dy = - (1/51) * dt

we integrate on both sides:

ln (y -19) = - (1/51) * t + C

e ^ ln (y -19) = e ^ (- (1/51) * t + C)

y - 19 = C1 * e ^ (- (1/51) * t)

y = 19 + C1 * e ^ (- (1/51) * t)

replacing, and (0) = 95

95 = 19 + C1 * e ^ (- (1/51) * 0)

95 = 19 + C1 * e ^ 0

95 = 19 + C1

C1 = 95 - 19 = 76

Therefore the equation would be:

y (t) = 19 + 76 * e ^ (- (1/51) * t)

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.